GIFT  OF 


mi 

^.  ,//<« 


F, 


A  TREATISE 


ON 


ELEMENTARY  GEOMETRY 


WITH  APPENDICES  CONTAINING 

A    COLLECTION    OF    EXEKCISES    FOR    STUDENTS 

AND 

AN  INTRODUCTION   TO  MODERN  GEOMETRY. 


BY 


WILLIAM  CHAUVENET,  LL.D., 

PROFESSOR  OP  MATHEMATICS  AND  ASTRONOMY  IN  WASHINGTON  UNIVERSITY. 


PHILADELPHIA      - 
J.  B.  LIPPINCOTT    &    CO. 

mo. 


Entered  according  to  Act  of  Congress,  in  the  year  1870,  by 
).  B.  LIPPINCOTT   &  CO., 

In  tne  Clerk's  Office  of  the  District  Court  of  the  United  States  for  the  Eastern  District  of 

Pennsylvania. 


IIPPINCOTT'S    PRESS, 


C$11 
PREFACE, 


THE  invention  of  Analytic  Geometry  by  DESCARTES  in  the 
early  part  of  the  seventeenth  century,  quickly  followed  by  that 
of  the  Infinitesimal  Calculus  by  NEWTON  and  LEIBNITZ,  pro 
duced  a  complete  revolution  in  the  mathematical  sciences  them 
selves  and  accelerated  in  an  astonishing  degree  the  progress  of 
all  the  sciences  in  which  mathematics  are  applied,  but  arrested 
for  a  time  the  progress  of  pure  geometry.  The  new  methods, 
characterized  by  great  generality  and  facility  in  their  application 
to  problems  of  the  most  varied  kinds,  offered  to  the  succeeding 
generations  of  investigators  more  inviting  fields  of  research  and 
promises  of  surer  and  richer  reward  than  the  special  and  ap 
parently  more  restricted  methods  of  the  ancients.  During  the 
eighteenth  century  hardly  any  important  addition  to  geometry 
was  made  that  was  not  the  direct  product,  either  of  the  Cartesian 
method  alone,  or  of  that  method  in  alliance  with  the  Infinitesi 
mal  Calculus. 

With  the  present  century,  however,  a  new  era  commenced  in 
pure  geometry.  The  first  impulse  was  given  by  the  Descriptive 
Geometry  of  MONGE  ;  then  followed  CARNOT'S  Theory  of  Trans 
versals,  PONCELET'S  Projective  Properties  of  Figures  and  Method 
of  Reciprocal  Polars,  the  researches  of  STEINER,  POINSOT,  GER- 
GONNE,  CAYLEY,  MACCULLAGH,  and  many  others,  crowned  by 
the  brilliant  discoveries  of  CHASLES. 

All  this  progress,  it  is  true,  has  been  chiefly  in  the  higher 
departments  of  pure  geometry,  and  has  not  yet  essentially  changed 

3G3802  s 


4  PREFACE. 

the  subsianee  or  fortft  of  what  is  known  as  Elementary  Geometry, 
which  is  little  more  than  the  Geometry  of  EUCLID  in  a  modern 
dress,  with  certain  necessary  additions  in  solid  geometry;  for, 
although  some  of  the  recent  discoveries  are  of  a  remarkably 
simple  character  and  (if  simplicity  were  the  only  requisite)  might 
be  introduced  into  the  elements,  it  is  generally  conceded  that  in 
elementary  instruction  it  is  most  expedient  to  commence  with  the 
Euclidian  geometry,  and  to  reserve  the  new  developments  for 
subsequent  study  under  the  name  of  the  Modern  Geometry. 

Nevertheless,  this  advance  in  the  general  science  has  not  failed 
to  produce  its  legitimate  effect  upon  the  primary  branch;  and 
the  modern  treatises  on  the  elements,  especially  in  France,  from 
that  of  LEGENDRE  in  1794  to  that  of  KOTJCHE  and  COMBEROUSSE 
in  1868,  exhibit  a  gradual  and  marked  improvement  both  in 
matter  and  method. 

In  the  following  treatise,  designed  especially  for  use  in  colleges 
and  schools,  I  have  endeavored  to  set  forth  the  elements  with  all 
the  rigor  and  completeness  demanded  by  the  present  state  of  the 
general  science,  without  seriously  departing  from  the  established 
order  of  the  propositions,  or  sacrificing  the  simplicity  of  demon 
stration  required  in  a  purely  elementary  work.  Some  subjects, 
not  usually  included  in  elementary  works,  are  so  placed  that  they 
may  be  omitted  without  breaking  the  chain  of  demonstration, 
and  the  remainder  may  be  used  as  an  abridged  course  in  those 
schools  where  the  time  allotted  to  the  study  does  liot  suffice  for 
the  perusal  of  the  whole.  Such,  for  example,  are  the  articles  on 
Maxima  and  Minima  at  the  end  of  Book  V.  and  those  on  Similar 
Polyedrons  and  the  Regular  Polyedrons  at  the  end  of  Book  VII. 

As  the  student  can  make  no  solid  acquisitions  in  geometry 
without  frequent  practice  in  the  application  of  the  principles  he 
has  acquired,  a  copious  collection  of  exercises  is  given  in  the 
Appendix.  The  discouraging  difficulties  which  the  young  student 
commonly  experiences  in  his  first  attempts  at  demonstrating  new 
theorems,  or  solving  new  problems,  are  here  obviated  in  a  great 


PREFACE.  5 

degree  by  giving  him  such  suggestions  for  the  solution  of  many 
of  the  exercises  as  may  fairly  be  presumed  to  be  necessary  for 
him  at  the  successive  stages  of  his  progress.  These  suggestions 
are  given  with  less  and  less  frequency  as  he  advances,  and  he 
is  finally  left  to  rely  entirely  upon  his  own  resources  when  he 
may  be  supposed  to  have  acquired  by  practice  considerable 
familiarity  with  principles,  and  dexterity  in  their  application. 

The  Appendix  on  the  Modern  Geometry,  although  restricted  to 
the  properties  of  the  straight  line  and  circle,  will  serve  a  good 
purpose,  it  is  hoped,  either  as  an  introduction  to  such  works  as 
those  of  PONCELET  and  CHASLES  in  which  the  methods  of  pure 
geometry  are  employed,  or  as  a  companion  to  the  works  of 
SALMON  and  others  in  which  the  new  geometry  is  treated  by  the 
analytic  method. 

In  the  preparation  of  this  work,  I  have  derived  valuable  aid 
from  a  number  of  the  more  recent  French  treatises  on  Element 
ary  Geometry,  and  especially  from  those  of  BOBILLIER,  BRIOT, 
COMPAGNON,  LEGENDKE  (edited  by  Blanchet),  and  the  very 
complete  Traite  de  Geometrie  filementaire  of  ROUCHE  and  COM- 
BEKOUSSE.  The  last  named  work  has  furnished  many  of  the 
exercises  of  Appendix  I.  and  much  of  the  matter  of  Appendix  II. 

WASHINGTON  UNIVERSITY, 

ST.  Louis,  June  1,  1869. 
1* 


CONTENTS 


INTRODUCTION 


PLANE  GEOMETKY. 

BOOK  I. 
RECTILINEAR  FIGURES 12 

BOOK  II. 
THE  CIRCLE °2 

BOOK  III. 
PROPORTIONAL  LINES.    SIMILAR  FIGURES 91 

BOOK  IV. 

COMPARISON  AND  MEASUREMENT  OF  THE  SURFACES  OF  EECTILINEAR 
FIGURES 126 

BOOK  V. 

REGULAR  POLYGONS.     MEASUREMENT  OF  THE  CIRCLE.    MAXIMA  AND 
MINIMA  OF  PLANE  FIGURES 142 


GEOMETRY  OF  SPACE. 

BOOK  VI. 

THE  PLANE.    POLYEDRAL  ANGLES 171 

7 


8  CONTENTS. 

BOOK  VII. 

PAGE 
POLYEDRONS 196 

BOOK  VIII. 

THE  THREE   BOUND   BODIES.     THE  CYLINDER.     THE  CONE.     THE 
SPHERE 238 

BOOK  IX. 
MEASUREMENT  OF  THE  THREE  BOUND  BODIES 271 


APPENDIX  I. 
EXERCISES  IN  ELEMENTARY  GEOMETRY 291 

APPENDIX  II. 

INTRODUCTION    TO    MODERN   GEOMETRY 333 


ELEMENTS  OF  GEOMETRY. 


INTRODUCTION. 

1.  EVERY  person  possesses  a  conception  of  space  indefinitely  ex 
tended  in  all  directions.     Material  bodies  occupy  finite,  or  limited, 
portions  of  space.     The  portion  of  space  which  a  body  occupies  can 
be  conceived  as  abstracted  from  the  matter  of  which  the  body  is 
composed,  and  is  called  a  geometrical  solid.     The  material  body  filling 
the  space  is  called  a  physical  solid.     A  geometrical  solid  is,  therefore, 
merely  the  form,  or  figure,  of  a  physical  solid.     In  this  work,  since 
only  geometrical  solids  will  be  considered,  we  shall,  for  brevity,  call 
them  simply  solids. 

2.  Definitions.  In  geometry,  then,  a  solid  is  a  limited,  or  bounded, 
portion  of  space. 

The  limits,  or  boundaries,  of  a  solid  are  surfaces. 
The  limits,  or  boundaries,  of  a  surface  are  lines. 
The  limits  of  a  line  are  points. 

3.  A  solid  has  extension  in  all  directions ;  but  for  the  purpose  of 
measuring  its  magnitude,  it  is  considered  as  having  three  specific 
dimensions,  called  length,  breadth  and  thickness. 

A  surface  has  only  two  dimensions,  length  and  breadth. 

A  line  has  only  one  dimension,  namely,  length.  The  intersection 
of  two  surfaces  is  a  line. 

A  point  has  no  extension,  and  therefore  neither  length,  breadth 
nor  thickness.  The  intersection  of  two  lines  is  a  point. 

4.  Although  our  first  notion  of  a  surface,  as  expressed  in  the  defi 
nition  above  given,  is  that  of  the  boundary  of  a  solid,  we  can  suppose 


10  G  BO  MET  BY. 

such  boundary  to  be  abstracted  and  considered  separately  from  the 
solid.  Moreover,  we  may  suppose  a  surface  of  indefinite  extent  as  to 
length  and  breadth ;  such  a  surface  has  no  limits. 

Similarly,  a  line  may  be  considered,  not  only  as  the  limit  of  a 
surface,  but  as  abstracted  from  the  surface  and  existing  separately  in 
space.  Moreover,  we  may  suppose  a  line  of  indefinite  length,  or 
without  limits. 

Finally,  a  point  may  be  considered,  not  merely  as  a  limit  of  a  line, 
but  abstractly  as  having  only  position  in  space. 

5.  Definitions.  A  straight  line  is  the  shortest          A 

line  between  two  points  ;  as  AB. 

Since  our  first  conception  of  a  straight  line  may  be  regarded  as 
derived  from  a  comparison  of  all  the  lines  that  can  be  imagined  to 
exist  between  two  points,  i.e.,  of  lines  of  limited  length,  this  definition 
(which  is  the  most  common  one)  may  be  admitted  as  expressing  such 
a  first  conception ;  but  since  we  can  suppose  straight  lines  of  indefi 
nite  extent,  a  more  general  definition  is  the  following : 

A  straight  line  is  a  line  of  which  every  portion  is  the  shortest  line 
between  the  points  limiting  that  portion. 

A  broken  line  is  a  line  composed  of  differ 
ent  successive  straight  lines;  as  ABCDEF. 

A  curved  line,  or  simply  a  curve,  is  a  line        ^ -^  c 

no  portion  of  which  is  straight;  as  ABC.  A 

If  a  point  moves  along  a  line,  it  is  said  to  describe  the  line. 

6.  Definitions.  A  plane  surface,  or  simply  a 
plane,  is  a  surface  in  which,  if  any  two  points 
are  taken,   the  straight    line  joining  these 
points  lies  wholly  in  the  surface. 

A  curved  surface  is  a  surface  no  portion  of  which  is  plane. 

7.  Solids  are  classified  according  to  the  nature  of  the  surfaces 
which  limit  them.     The  most  simple  are  bounded  by  planes. 

8.  Definitions.  A  geometrical  figure  is  any  combination  of  points, 
lines,  surfaces,  or  solids,  formed  under  given  conditions.     Figures 
formed  by  points  and  lines  in  a  plane  are  called  plane  figures.     Those 
formed  by  straight  lines  alone  are  called  rectilinear,  or  right-lined, 
figures  ;  a  straight  line  being  often  called  a  right  line. 


INTRODUCTION.  11 

9.  Definitions.   Geometry  may  be  defined  as  the  science  of  extension 
and  position.     More  specifically,  it  is  the  science  which  treats  of  the 
construction  of  figures  under  given  conditions,  of  their  measurement, 
and  of  their  properties. 

Plane  geometry  treats  of  plane  figures. 

The  consideration  of  all  other  figures  belongs  to  the  geometry  of 
space,  also  called  the  geometry  of  three  dimensions. 

10.  Some  terms  of  frequent  use  in  geometry  are  here  defined. 

A  theorem  is  a  truth  requiring  demonstration.  A  lemma  is  an 
auxiliary  theorem  employed  in  the  demonstration  of  another  theo 
rem.  A  problem  is  a  question  proposed  for  solution.  An  axiom  is  a 
truth  assumed  as  self-evident.  A  postulate  (in  geometry)  assumes 
the  possibility  of  the  solution  of  some  problem. 

Theorems,  problems,  axioms  and  postulates  are  all  called  propo 
sitions. 

A,  corollary  is  an  immediate  consequence  deduced  from  one  or  more 
propositions.  A  scholium  is  a  remark  upon  one  or  more  propositions, 
pointing  out  their  use,  their  connection,  their  limitation,  or  their 
extension.  An  hypothesis  is  a  supposition,  made  either  in  the  enun 
ciation  of  a  proposition,  or  in  the  course  of  a  demonstration. 


PLANE  GEOMETRY. 


BOOK  I. 

RECTILINEAR  FIGURES. 
THE  STEAIGHT  LINE. 

1.  AXIOM.  There  can  be  but  one  straight  line  between  the  same 
two  points. 

2.  Postulate.  A  straight  line  can  be  drawn  between  any  two  points ; 
and  any  straight  line  can  be  produced  (i.  e.,  prolonged)  indefinitely. 

3.  Axiom.  If  two  indefinite  straight  lines  coincide  in  two  points, 
they  coincide  throughout  their  whole  extent,  and  form  but  one  line. 

Hence  two  points  determine  a  straight  line;  and  a  straight  line 
may  be  designated  by  any  two  of  its  points. 

4.  Different  straight  lines  drawn  from  the  same  point  are  said  to 
have  different  directions ;  as  OA,  OD,  etc.     The  c 
point  from  which  they  are  drawn,  or  at  which 

they  commence,  is  often  called  the  origin. 

If  any  one  of  the  lines,  as  OA,  be  produced        .J 

through  0,  the  portions  OA,  OB,  on  opposite 

sides  of  0,  may  be  regarded  as  two  different  lines  having  opposite 

directions  reckoned  from  the  common  origin  0. 

Hence,  also,  every  straight  line  AB  has  two  opposite  directions, 

namely,  from  A  toward  B  (A  being  regarded  as  

its  origin)  expressed  by  AB,  and  from  B  toward 

A  (B  being  regarded  as  its  origin)  expressed  by  BA.     If  a  line  AB 

is  to  be  produced  through  B,  that  is,  toward 

C,  we  should  express  this   by  saying  that        -     +  +     - 

AB  is  to  be  produced;    but  if  it  is  to  be 

12 


BOOK     I.  13 

produced  through  A,  that  is  toward  D,  we  should  express  this  by 
saying  that  BA  is  to  be  produced. 

ANGLES. 

5.  Definition.  An  angle  is  a  figure  formed  by  two  B 
straight   lines   drawn    from    the   same   point;    thus 

OA,  OB  form  an  angle  at  0.  The  lines  OA,  OB 
are  called  the  sides  of  the  angle ;  the  common  point 
0,  its  vertex. 

An  isolated  angle  may  be  designated  by  the  letter  at  its  vertex,  as 
"the  angle  0 ;"  but  when  several  angles  are  formed  at  the  same 
point  by  different  lines,  as  OA,  OB,  00,  we  desig- 

/> 

nate  the  angle  intended  by  three  letters ;  namely,  by 
one  letter  on  each  of  its  sides,  together  with  the  one 
at  its  vertex,  which  must  be  written  between  the  other 
two.  Thus,  with  these  lines  there  are  formed  three 
different  angles,  which  are  distinguished  as  A  OB,  BOC  and  AOC. 

Two  angles,  such  as  A  OB,  BOO,  which  have  the  same  vertex  0 
and  a  common  side  OB  between  them,  are  called  adjacent. 

6.  Definition.  Two  angles  are  equal  when  one  can  be  placed  upon 
the  other  so  that  they  shall  coincide.     Thus,  the 

angles  A  OB  and  A'  0' B'  are  equal,  if  A'  0' Br  can 
be  superposed  upon  A  OB  so  that  while  0'  A  coin 
cides  with  OA,  0' B'  shall  also  coincide  with  OB. 
The  equality  of  the  two  angles  is  not  affected  by 
producing  the  sides ;  for  the  coincident  sides  con 
tinue  to  coincide  when  produced  indefinitely  (3).*  Thus  the  magnitude 
of  an  angle  is  independent  of  the  length  of  its  sides. 

7.  A  clear  notion  of  the  magnitude  of  an  angle  will  be  obtained 
by  supposing  that  one  of  its  sides,  as  OB,  was  at  first 

coincident  with  the  other  side  OA,  and  that  it  has 

revolved  about  the  point  0  (turning  upon  0  as  the  leg 

of  a  pair  of  dividers  turns  upon  its  hinge)  until  it 

has  arrived  at  the  position  OB.     During  this  revolution  the  movable 

side  makes  with  the  fixed  side  a  varying  angle,  which  increases  by 

insensible  degrees,  that  is,  continuously;  and  the  revolving  line  is 

*  An  Arabic  numeral  alone  refers  to  an  article  in  the  same  Book ;  but  in  refer 
ring  to  articles  in  another  Book,  the  number  of  the  Book  is  also  given. 

2 


14  GEOMETRY. 

said  to  describe,  or  to  generate,  the  angle  A  OB.     By  continuing  the 
revolution,  an  angle  of  any  magnitude  may  be  generated. 

It  is  evident  from  this  mode  of  generation,  as  well  as  from  the  defi 
nition  (6),  that  the  magnitude  of  an  angle  is  independent  of  the 
length  of  its  sides. 

PEKPENDICULAKS  AND  OBLIQUE  LINES. 

8.  Definition.  When  one  straight  line  meets  another,  so  as  to  make 
two  adjacent  angles  equal,   each  of  these  angles  is  called  a  right 
angle ;  and  the  first  line  is  said  to  be  perpendicular  to  the  second. 

Thus,  if  A OC  and  BOG  are  equal  angles,  c 

each  is  a  right  angle,  and  the  line  CO  is  per 
pendicular  to  AB. 

Intersecting  lines  not  perpendicular  are  said 
to  be  oblique  to  each  other. 

PEOPOSITION  I.— THEOKEM. 

9.  At  a  given  point  in  a  straight  line  one  perpendicular  to  the  line 
can  be  drawn,  and  but  one. 

Let  0  be  the  given  point  in  the  line  AB.  Suppose  a  line  01), 
constantly  passing  through  0,  to  revolve  about  G 

0,  starting  from  the  position  OA.  In  any  one 
of  its  successive  positions,  it  makes  two  different 
angles  with  the  line  AB;  one,  AOD,  with  the 
portion  OA;  and  another,  BOD,  with  the  por 
tion  OB.  As  it  revolves  from  the  position  OA  around  to  the  posi 
tion  OB,  the  angle  A  OD  will  continuously  increase,,  and  the  angle 
BOD  will  continuously  decrease.  There  will  therefore  be  one  posi 
tion,  as  OC,  where  the  two  angles  become  equal;  and  there  can  evi 
dently  be  but  one. 

10.  Corollary.   All   right   angles   are   equal.     That   is,  the   right 
angles  AOC,  BOG  made  by  a  line  CO 

meeting  AB,  are  each  equal  to  each  of 
the  right  angles  A' O'C",   B'O'C',  made 
by  a  line  C'  0'" meeting  any  other  line 
A'B'.     For,  the  line  A'O'B'  can  be  ap-       B~- 
plied  to  the  line  A  OB,  so  that  0'  shall 


Bt          Of 


BOOK      I.  15 

fall  upon  0,  and  then  0'Cf  will  fall  upon  OC,  unless  there  can  be 
two  perpendiculars  to  AB  at  0,  which  by  the  preceding  proposition 
is  impossible.  The  lines  will  therefore  coincide  and  the  angles  will 
be  equal  (6). 

PEOPOSITION  II.— THEOREM. 

11.  The  two  adjacent  angles  which  one  straight   line  makes  with 
another  are  together  equal  to  two  right  angles. 

If  the  two  angles  are  equal,  they  are  right  angles  by  the  definition 
(8),  and  no  proof  is  necessary. 

If  they  are  not  equal,  as  A  OD  and  B  OD,  still  the  sum  of  A  OD 
and  BOD  is  equal  to  two  right  angles.  For,  let  OC  be  drawn  at  0 
perpendicular  to  AB.  The  angle  AOD  is  the  c 

sum  of  the  two  angles  A  OC  and  COD.     Adding 
the  angle  B  OD,  the  sum  of  the  two  angles  A  OD 
and  BOD  is  the  sum  of  the  three  angles  AOC, 
COD  and  BOD.     The  first  of  these  three  is  a       B 
right  angle,  and  the  other  two  are  together  equal  to  the  right  angle 
BOG;  hence  the  sum  of  the  angles  AOD  and  BOD  is  equal  to  two 
right  angles. 

12.  Corollary  I.  If  one   of  the  two  adjacent   angles  which  one 
line  makes  with  another  is  a  right  angle,  the  other  is  also  a  right 
angle. 

n 

13.  Corollary  II.  If  a  line   CD  is  perpen 
dicular  to  another  line  AB,  then,  reciprocally, 
the  line  AB  is  perpendicular  to  CD.     For, 
CO  being  perpendicular  to  AB  at  0,  AOC    E 
is  a  right  angle,  hence  (Cor.  I.)  AOD  is  a 
right  angle,  and  A  0  or  AB  is  perpendicular 

to  CD. 

14.  Corollary  III.  The  sum  of  all  the  consecutive  angles,  A  OB, 
BOC,  COD,  DOE,  formed  on  the  same  side 

of  a  straight  line  AE,  at  a  common  point  0, 
is  equal  to  two  right  angles.  For,  their  sum 
is  equal  to  the  sum  of  the  two  adjacent  angles 
A  OB,  B  OE,  which  by  the  proposition  is  equal 
to  two  right  angles. 


16 


GEOMETRY. 


15.  Corollary  TV.  The  sum  of  all  the  consecutive  angles  AOB, 
BOC,    COD,   DOE,   EOA,   formed   about   a 

point  0,  is  equal  to  four  right  angles.  For, 
if  two  straight  lines  are  drawn  through  0, 
perpendicular  to  each  other,  the  sum  of  all 
the  consecutive  angles  formed  about  0  will 
be  equal  to  the  four  right  angles  formed  by 
the  perpendiculars. 

16.  Scholium.  A  straight  line  revolving   from  the  position   OA 
around  to   the  position   OB  describes  the  two  c 

right  angles  AOC  and  COB;  hence  OA  and 
OB,  regarded  as  two  different  lines  having 
opposite  directions  (4),  are  frequently  said  to 
make  an  angle  with  each  other  equal  to  two 
right  angles. 

A  line  revolving  from  the  position  OA  from  right  to  left,  that  is, 
successively  into  the  positions  OC,  OB,  OD, 
when  it  has  arrived  at  the  position  OD  will 
have  described  an  angle  greater  than  two 
right  angles.  On  the  other  hand,  if  the 
position  OD  is  reached  by  revolving  from 
left  to  right,  that  is,  successively  into  the 
positions  OE,  OD,  then  the  angle  AOD  is 
less  than  two  right  angles.  Thus,  any  two 

straight  lines  drawn  from  a  common  point  make  two  different  angles 
with  each  other,  one  less  and  the  other  greater  than  two  right  angles. 
Hereafter  the  angle  which  is  less  than  two  right  angles  will  be 
understood,  unless  otherwise  expressly  stated. 

17.  Definitions.     An  acute  angle  is  an  angle  c 
less  than  a  right  angle;  as  AOD.     An  obtuse 

angle  is  an  angle  greater  than  a  right  angle;  as 
BOD. 

18.  When  the  sum  of  two  angles  is  equal  to  a 

right  angle,  each  is  called  the  complement  of  the  other.  Thus  DOC 
is  the  complement  of  AOD,  and  AOD  is  the  complement  of  DOC. 

19.  When  the  sum  of  two  angles  is  equal  to  two  right  angles,  each 
is  called  the  supplement  of  the  other.     Thus  BOD  is  the  supplement 
of  AOD,  and  AOD  is  the  supplement  of  BOD. 


BOOK     I.  17 

20.  It  is  evident  that  the  complements  of  equal  angles  are  equal 
to  each  other;  and  also  that  the  supplements  of  equal  angles  are 
equal  to  each  other. 

PROPOSITION  III.— THEOKEM. 

21.  Conversely,  if  the  sum  of  two  adjacent  angles  is  equal  to  two 
right  angles,  their  exterior  sides  are  in  the  same  straight  line. 

Let  the  sum  of  the  adjacent  angles  AOD,                                D 
B  OD,  be  equal  to  two  right  angles  ;  then,  OA                            .x' 
and  OB  are  in  the  same  straight  line.  -^    

For  BOD  is  the  supplement  of  AOD  (19), 

and  is  therefore  identical  with  the  angle  which  OD  makes  with  the 
prolongation  of  AO  (11).  Therefore  OB  and  the  prolongation  of 
A  0  are  the  same  line. 

22.  Every  proposition  consists  of  an  hypothesis  and  a  conclusion. 
The  converse  of  a  proposition  is  a  second  proposition  of  which  the 
hypothesis  and  conclusion  are  respectively  the  conclusion  and  hy 
pothesis  of  the  first.     For  example,  Proposition  II.  may  be  enun 
ciated  thus : 

Hypothesis — if  two  adjacent  angles  have  their  exterior  sides  in  the 
same  straight  line,  then — Conclusion — the  sum  of  these  adjacent 
angles  is  equal  to  two  right  angles. 

And  Proposition  III.  may  be  enunciated  thus : 

Hypothesis — if  the  sum  of  two  adjacent  angles  is  equal  to  two 
right  angles,  then — Conclusion — these  adjacent  angles  have  their 
exterior  sides  in  the  same  straight  line. 

Each  of  these  propositions  is  therefore  the  converse  of  the  other. 

A  proposition  and  its  converse  are  however  not  always  both  true. 

PEOPOSITION  IV.— THEOREM. 

23.  If  two  straight  lines  intersect  each  other,  the  opposite  (or  vertical) 
angles  are  equal. 

Let  AB  and  CD  intersect  in  0 ;  then  will  the 
opposite,  or  vertical,  angles  A  0  C  and  B  OD  be 
equal.  For,  each  of  these  angles  is  the  supple 
ment  of  the  same  angle  BOC,  or  AOD,  and 
hence  they  are  equal  (20). 

2  *  B 


18 


GEOMETRY. 


In  like  manner  it  is  proved  that  the  opposite  angles  A  OD  and 
BOG  are  equal. 

24.  Corollary  I.  The  straight  line  EOF  which  bisects  the  angle 
AOC  also  bisects  its  vertical  angle  BOD.     For,  the  angle  FOD  is 
equal  to  its  vertical  angle  EOC,  and  FOB  is 

equal  to  its  vertical  angle  EOA ;  therefore  if 
EOC  and  EOA  are  equal,  FOD  and  FOB 
are  equal. 

25.  Corollary  II.   The  two   straight   lines 
EOF,  HOG,  which  bisect  the  two  pairs  of 

vertical    angles,  are    perpendicular   to   each  ! 

other.     For,  HOC  =  HOB  and   COE  = 

BOF-,  hence,  by  addition,  HOC  +  COE  =  HOB  +  BOF;  that 
is,  HOE  =  HOF;  therefore,  by  the  definition  (8),  HO  is  perpen 
dicular  to  FE. 


\ 

E 


PROPOSITION  V.-THEOREM. 

X      26.  From  a  given  point  without  a  straight  line,  one  perpendicular 
can  be  drawn  to  that  line,  and  but  one. 

Let  AB  be  the  given  straight  line  and  P  the  given  point. 

The  line  AB  divides  the  plane  in  which  it 
is  situated  into  two  portions.  Let  the  por 
tion  containing  P,  which  we  suppose  to  be 
the  upper  portion,  be  revolved  about  the  line 
AB  (i.e.,  folded  over)  until  the  point  P  comes 
into  the  lower  portion;  and  let  P'  be  that 
point  in  the  plane  with  which  P  coincides 
after  this  revolution.  Restoring  P  to  its 

original  position,  join  PPf,  cutting  AB  in  C,  and  again  revolve  the 
upper  portion  of  the  plane  about  AB  until  P  again  coincides  with 
P'.  Since  the  line  AB  is  fixed  during  the  revolution,  the  point  C  is 
fixed;  therefore  PC  will  coincide  with  P'C,  and  the  angle  PCD 
with  the  angle  P' CD.  These  angles  are  therefore  equal  (6),  and 
BC  is  perpendicular  to  PP'  (8),  or  PC  perpendicular  to  AB  (13). 
There  can  therefore  be  one  perpendicular  from  the  point  P  to  the 
line  AB. 

Moreover,  PC  is  the  only  perpendicular.     Let  PD  be  any  other 


BOOK    I.  19 

line  drawn  from  P  to  AB,  and  join  P'D.  Then,  when  the  upper 
portion  of  the  plane  is  revolved  until  P  coincides  with  P',  D  being 
fixed,  PD  coincides  with  P'D,  and  consequently  the  angle  PDC  with 
the  angle  P'DC.  Hence  the  angles  PDC  and  P'DC  are  equal. 
Now  PP'  being  the  only  straight  line  that  can  be  drawn  from  P  to 
P'  (1),  PDP'  is  not  a  straight  line ;  and  if  PD  is  produced  to  E, 
PDE  and  DP'  are  different  straight  lines.  Hence  the  angle  PDP' 
is  less  than  two  right  angles,  and  its  half,  PDC,  is  less  than  one 
right  angle ;  that  is,  PD  is  an  oblique  line.  Therefore  PC  is  the 
only  perpendicular. 

27.  Corollary.  Of  the  two  angles  which  any  oblique  line  drawn 
from  P  makes  with  AB,  that  one  is  acute  within  which  the  perpen 
dicular  from  Pupon  AB  falls;  thus,  PDC  is  acute. 

PKOPOSITION  VI.— THEOKEM. 

28.  The  perpendicular  is  the  shortest  line  that  can  be  drawn  from  a 
point  to  a  straight  line. 

Let  PC  be  the  perpendicular,  and  PD  any  oblique  line,  from  the 
point  P  to  the  line  AB.    Then  PC  <  PD. 

For,  produce  PC  to  P',  making  OP'  = 
CP,  and  join  P'D.  When  the  portion  of  the 
plane  which  contains  P  is  revolved  about 
AB,  as  in  the  preceding  proposition,  until  P  A~~ 
coincides  with  P',  PD  also  coincides  with 
P'D',  and  hence  PD  =  P'D.  But  the 
straight  line  PP',  being  the  shortest  distance 
between  the  points  P  and  P',  is  less  than  the  broken  line  PDP'. 
Therefore  PC,  the  half  of  the  straight  line,  is  less  than  PD,  the  half 
of  the  broken  line. 

29.  Definition.   By  the  distance  of  a  point  from  a  line  is  always 
understood   the  shortest  distance.      By   the  preceding   proposition, 
therefore,  the  perpendicular  measures  the  distance  of  a  point  from  a 
straight  line. 

Also,  by  the  distance  of  one  point  from  another  is  understood  the 
shortest  distance,  that  is,  the  straight  line  between  the  points. 


20 


GEOMETRY. 


A      D 


E      B 


PKOPOSITION  VII.— THEOKEM. 

30.  Two  oblique  lines  drawn  from  the  same  point  to  a  straight  line, 
cutting  off  equal  distances  from  the  foot  of  the  perpendicular,  are  equal. 

Let  the  oblique  lines  PD,  PE,  meet  the  line  AB  in  the  points  D 
and  E,  cutting  off  the  equal  distances  CD 
and  CE  from  the  foot  of  the  perpendicular. 
Then  PD  =  PE. 

For,  DCE  being  perpendicular  to  -PC, 
and  CD  =  CE,  the  figure  PCD  may  be  re 
volved  about  PC  into  coincidence  with 

PCE;  and  since  the  point  D  will  fall  on  E,  PD  will  coincide  with 
PE.     Therefore  PD  =  PE. 

31.  Corollary.   The  angles  PDC  and  PEC  are  equal ;  that  is,  two 
equal  straight  lines  from  a  point  to  a  straight  line  make  equal  acute 
angles  with  that  line. 

32.  Definition.  A  broken  line,  as  ABCDE,  is  called  convex,  when 
no  one  of  its  component  straight  lines,  if  pro 
duced,  can   enter  the   space  enclosed   by  the 

broken  line  and  the  straight  line  joining  its 
extremities. 


PBOPOSITION  VIIL— THEOEEM. 

33.  A  convex  broken  line  is  less  than  any  other  line  which  envelops  it 
and  has  the  same  extremities. 

Let  the  convex  broken  line  AFGE  have  the  same  extremities  A, 
E,  as  the  line  ABCDE,  and  be  enveloped  by 
it ;  that  is,  wholly  included  within  the  space 
bounded  by  ABCDE  and  the  straight  line 
AK     Then  AFGE  <  AB  CDK 

For,  produce  AF  and  FG  to  meet  the  en 
veloping  line  in  H  and  K.  Imagine  ABCDE  to  be  the  path  of  a 
point  moving  from  A  to  E.  If  the  straight  line  AH  be  substituted 
for  ABC,  the  path  AHDE  will  be  shorter  than  the  path  ABCDE, 
the  portion  HDE  being  common  to  both.  If,  further,  the  straight 
line  FK  be  substituted  for  FHDK,  the  path  AFKE  will  be  a  still 
shorter  path  from  A  to  E.  And  if,  finally,  GE  be  substituted  for 


BOOK     I.  21 

GKE,  AFGE  will  be  a  still  shorter  path.     Therefore,  AFGE  is  less 
than  any  enveloping  line. 

34.  Scholium.  The  preceding  demonstration  applies  when  the  en 
veloping  line  is  a  curve,  or  any  species  of  line  whatever. 


PKOPOSITION  IX— THEOKEM. 

35.  Of  tivo  oblique  lines  drawn  from  the  same  point  to  the  same 
straight  line,  that  is  the  greater  which  cuts  of  upon  the  line  the  greater 
distance  from  the  perpendicular. 

Let  PC  be  the  perpendicular  from  P  to  AB,  and  suppose  CE  > 
CD;  thenPE>PD. 

For,  produce  PC  to  P',  making  CP'  = 
CP,  and  join  DP',  EP'.  Then,  as  in  Pro 
position  VI.,  we  have  PD  =  P'D,  and  PE 
=  P'E.  But  (33),  the  broken  line  PDP' 
is  less  than  the  enveloping  line  PEP'-, 
therefore  PD,  the  half  of  PDP',  is  less  than 
PE,  the  half  of  PEP'. 

If  the  two  oblique  lines  are  on  opposite  sides  of  the  perpendicular, 
as  PE  and  PD',  and  if  CE  >  CD',  take  CD  =  CD',  and  join  PD. 
Then,  as  above  PE  >  PD;  and,  by  Proposition  VII.,  PD  =  PD' ; 
hence  PE  >  PD'. 

36.  Corollary  I.  (Converse  of  Proposition  VII.).     Two  equal  ob 
lique  lines  cut  off  equal  distances  from  the  perpendicular. 

37.  Corollary  II.  (Converse  of  Proposition  IX.).     Of  two  unequal 
oblique  lines,  the  greater  cuts  off  the  greater  distance  from  the  per 
pendicular. 

PEOPOSITION  X.— THEOKEM. 

38.  If  a  perpendicular  is  erected  at  the  middle  of  a  straight  line, 
then, 

1st.  Every  point  in  the  perpendicular  is  equally  distant  from  the 
extremities  of  the  line ; 

2d.  Every  point  without  the  perpendicular  is  unequally  distant  from 
the  extremities  of  the  line. 


22 


GEOMETRY. 


Let  AB  be  a  finite  straight  line,  and  C its  middle  point;  then, 

1st.  Every  point  P  in  the  perpendicular 
erected  at  C  is  equally  distant  from  A  and  B. 
For,  since  CA  =  CB,  we  have  (30)  PA  =  PB. 

2d.  Any  point  Q  without  the  perpendicular 
is  unequally  distant  from  A  and  B.  For,  Q 
being  on  one  side  or  the  other  of  the  perpendicular,  one  of  the  lines 
QA,  QB  must  cut  the  perpendicular ;  let  it  be  QA  and  let  it  cut  in 
P;  join  PB.  The  straight  line  QB  is  less  than  the  broken  line 
QPB,  that  is,  QB  <  QP  -f  PB.  But  PB  =  PA]  therefore 
QB<QP  +  PA,  or  QB  <  QA. 

39.  Corollary.    Every  point  equally  distant  from  the  extremities 
of  a  straight  line  lies  in  the  perpendicular  erected  at  the  middle  of 
the  line. 

40.  Definition.    A  geometric  locus  is  the  assemblage  of  all  the 
points  which  possess  a  common  property. 

In  this  definition,  points  are  understood  to  have  a  common  property 
when  they  satisfy  the  same  geometrical  conditions. 

Thus,  since  all  the  points  in  the  perpendicular  erected  at  the 
middle  of  a  line  possess  the  common  property  of  being  equally  dis 
tant  from  the  extremities  of  the  line  (that  is,  satisfy  the  condition 
that  they  shall  be  equally  distant  from  those  extremities),  and  no 
other  points  possess  this  property,  the  perpendicular  is  the  locus  of 
these  points ;  so  that  the  preceding  proposition  and  its  corollary  are 
fully  covered  by  the  following  brief  statement: 

The  perpendicular  erected  at  the  middle  of  a  straight  line  is  the  locus 
of  all  the  points  which  are  equally  distant  from  the  extremities  of  that 
line.  rt 

41.  Scholium.  Two  points  are  sufficient  to  determine  a  straight 
line  (3)  ;  hence  any  two  points  each  of  which 

is  equally  distant  from  the  extremities  of  a 
straight  line  determine  the  perpendicular  at 
the  middle  of  the  line.  Thus  if  P  and  P' 
are  known  to  be  each  equally  distant  from 
A  and  B,  the  line  PP'  joining  these  points  is 
known  to  be  perpendicular  to  AB  at  its  mid 
dle  point. 


BOOK     I. 


23 


PAKALLEL  LINES. 

42.  Definition.  Parallel  lines  are  straight 
lines  which  lying  in  the  same  plane  cannot 
meet,  though  indefinitely  produced :  as  AB, 
CD. 


43.  Axiom.  Through  the  same  point  there  cannot  be  two  parallels 
to  the  same  straight  line. 

Thus,  if  through  a  point  P,  one  line  CD  is 
drawn  parallel  to  AB,  the  axiom  assumes 
that  any  other  line  drawn  through  P,  as 
EPF,  will  not  be  parallel  to  AB,  but  will 
meet  it,  if  both  EF  and  AB  be  sufficiently  produced. 

PKOPOSITION  XI.— THEOKEM. 

44.  Two  straight  lines  perpendicular  to  the  same  straight  line  are 
parallel. 

Let  AB  and  CD  be  perpendicular  to  AC]  then,  they  are  parallel. 

For,  if  they  could  meet  when  produced,  we 
should  have  from  one  point  (their  point  of 
meeting)  two  perpendiculars  to  the  same 
straight  line  AC,  which  (26)  is  impossible. 
Therefore  they  cannot  meet,  and  by  the  defi 
nition  (42)  are  parallel. 

45.  Corollary  I.    Through  a  given   point  a  parallel  to  a  given 
straight  line  can  always  be  drawn.     For,  let  C  be  the  given  point, 
and  AB  the  given  line.     From  C  a  perpendicular  CA  can  be  drawn 
to  AB  (26) ;  and  at  C  a  perpendicular  CD  to  CA  can  be  drawn  (9) ; 
and  by  the  preceding  proposition  CD  will  be  parallel  to  AB. 

46.  Corollary  II.  A  straight  line  perpendicular  to  one  of  two  par 
allels  is  perpendicular  to  the  other. 

Let  A  C  be  a  perpendicular  to  AB ;  it  will  also  be  perpendicular 
to  the  parallel  CD.  In  the  first  place  it  is  to  be  observed  that  A  C 
being  a  different  line  from  AB  cannot  also  be  parallel  to  CD  (43), 
and  must  therefore  meet  CD  in  some  point,  as  C.  Moreover  the 
perpendicular  to  AC  at  C  is  parallel  to  AB  (44)  and  must  coincide 
with  CD  (9)  and  (43).  Hence  AC  is  perpendicular  to  CD. 


24 


GEOMETRY. 


PROPOSITION  XII.— THEOREM. 

47.  Two  straight  lines  parallel  to  a  third  are  parallel  to  each  other. 
Let  CD  and  EF  be  parallel  to  AB\  then, 

they  are  parallel  to  each  other.     For,  if  they        

could  meet,  there  would  be  drawn  through  c                         D 

their  point  of  meeting  two  straight  lines  par-  E                          J> 
allel  to  the  same  straight  line,  which  (43)  is 
impossible.    Hence  they  cannot  meet,  and  are  parallel  to  each  other. 

48.  Definitions.     When  two  straight  lines  AB,  CD,  are  cut  by  a 
third  EF,  the  eight  angles  formed  at  their 

points  of  intersection  are  named  as  follows : 

The  four  angles,  1,  2,  3,  4,  without  the 
two  lines,  are  called  exterior  angles. 

The  four  angles,  5,  6,  7,  8,  within  the  two 
lines,  are  called  interior  angles. 

Two  exterior  angles  on  opposite  sides  of 
the  secant  line  and  not  adjacent — as  1,  3 — or  2,  4 — are  called  alter 
nate-exterior  angles. 

Two  interior  angles  on  opposite  sides  of  the  secant  line  and  not 
adjacent — as  5,  7 — or  6,  8 — are  called  alternate-interior  angles. 

Two  angles  similarly  situated  with  respect  both  to  the  secant  and 
to  the  line  intersected  by  it,  are  called  corresponding  angles;  as 
1,  5-^2,  64-3,  7-K  8. 


PROPOSITION  XIII.— THEOREM. 

49.  If  two  parallel  lines  are  cut  by  a  third  straight  line,  the  alternate- 
interior  angles  are  equal. 

Let  the  parallels  AB,  CD,  be  cut  by  the 
straight  line  EF  in  the  points  G  and  H; 
then,  the  alternate-interior  angles,  HGB 
and  GHC,  are  equal. 

For,  through  I,  the  middle  point  of  GH, 
suppose  the  indefinite  line  KIL  to  be  drawn 
perpendicular  to  AB;  it  will  also  (46)  be 
perpendicular  to  CD.  Conceive  the  por 
tion  IGB  of  the  figure,  including  the  per- 


BOOK     I.  25 

pendicular  IK,  to  be  revolved  in  its  own  plane  about  /  (as  upon  a 
pivot),  until  IG  comes  into  coincidence  with  its  equal  IH.  The 
angle  G IK  being  equal  to  its  vertical  angle  HIL,  the  indefinite  line 
IK  will  fall  upon  IL  and  form  with  it  but  one  line.  Moreover,  the 
point  G  being  then  at  H,  the  line  GB  which  is  perpendicular  to  IK 
will  then  coincide  with  HC  which  is  perpendicular  to  IL,  and  con 
sequently  the  angles  IGB  and  IHC  will  coincide.  Therefore  the 
angles  HGB  and  GHC  are  equal. 

Hence,  also,  their  supplements,  HGA  and  GHD,  are  equal. 

50.  Corollary  I.  The  alternate-exterior  angles,  AGE  and  DHF, 
being  equal  to  their  vertical  angles,  HGB  and  GHC,  are  also  equal 
to  each  other. 

51.  Corollary  II.  Any  one  of  the  eight  angles  is  equal  to  its  cor 
responding  angle.     Thus,  since  HGB  =  GHC  and  GHC  =FHD, 
there  follows  HGB  =  FHD;  etc. 

52.  Corollary  III.  The  sum  of  the  two  interior  angles  on  the  same 
side  of  the  secant  line  is  equal  to  two  right  angles.     For,  GHD  -j- 
HGB  =  GHD  -f  GHC  =  two  right  angles  (11). 

53.  Scholium.  When  the  secant  line  is  oblique  to  the  parallels, 
there  are  formed  four  equal  acute  angles  and  four  equal  obtuse 
angles,  and  each  acute  angle  is  the  supplement  of  each  obtuse  angle. 
But  if  any  one  of  the  eight  angles  is  a  right  angle,  they  are  all  right 
angles. 

PEOPOSITION  XIV.— THEOREM. 

54.  Conversely,  when  two  straight  lines  are  cut  by  a  third,  if  the  alter 
nate-interior  angles  are  equal,  these  two  straight  lines  are  parallel. 

Let  EF  cut  AB  and  CD  in  the  points  G  and  H,  and  let  HGB 
and  GHC  be  equal;  then,  AB  and  CD  are 
parallel. 

For,  a  parallel  to  AB  drawn  through  H 
makes  with  GH  an  interior  angle,  alternate 
to  HGB,  which  is  equal  to  HGB  (49); 
this  angle  must  therefore  coincide  with  the 
angle  GHC,  and  the  parallel  drawn  through 
H  must  coincide  with  CD.  That  is,  CD  is  parallel  to  AB. 

55.  Corollary  I.  If  the  alternate-exterior  angles  are  equal,  or  if  the 
corresponding  angles  are  equal,  the  two  lines  are  parallel. 

3 


26  GEOMETRY. 

56.  Corollary  II.   If  the  sum  of  the  two  interior  angles  on  the 
same  side  of  the  secant  line  is  equal  to  two  right  angles,  the  two 
lines  are  parallel. 

57.  Corollary  III.  From  (52)  and  (56)  it  follows  that,  when  two 
straight  lines  are  cut  by  a  third,  if  the  sum  of  two  interior  angles 
on  the  same  side  of  the  secant  line  is  not  two  right  angles,  the  two 
straight  lines  are  not  parallel ;  and  it  is  evident  that  they  will  meet, 
if  produced,  on  that  side  of  the  secant  line  on  which  the  two  in 
terior  angles  are  together  less  than  two  right  angles. 


PKOPOSITION  XV.— THEOKEM. 

58.  Two  parallels  are  everywhere  equally  distant. 

Let  AB  and  CD  be  two  indefinitely  extended  parallels ;  G  and  H 
any  two  points  in  CD',  GE  and  HF  the  per-          G          M          H 
pendiculars  from  G  and  H  upon  AB.     Then, 
GE  and  HF  are  also  perpendicular  to  CD     A 
(46),  and  measure  the  distance  between  the 

parallels  at  G  and  H,  or  at  E  and  F.     We  are  to  prove  that  GE  = 
HF. 

Let  M  be  the  middle  of  GH,  and  suppose  MN  drawn  perpendicu 
lar  to  GH  and  consequently  also  to  EF.  The  portion  of  the  figure 
on  the  right  of  MN m&y  be  revolved  upon  the  line  MN  (i.e.,  folded 
over)  ;  the  angles  at  M  and  N  being  right  angles  the  indefinite  lines 
MD  and  NB  will  fall  upon  MC  and  NA;  and  since  MH  =  MG, 
the  point  -ffwill  fall  upon  G,  so  that  HF  and  M€h  (being  then  per 
pendiculars  from  the  same  point  G  upon  the  same  straight  line  NA), 
will  coincide  (26).  Therefore  GE  =  HF. 

59.  Corollary.    The  lovus  (40)   of  all  the  M 
points  at  a  given  distance,  MN,  from  a  given 

straight  line  AB,  consists  of  two  parallel  A 
lines,  CD  and  C'D',  drawn  on  opposite  sides  cl 
of  AB,  at  the  given  distance  from  it. 


BOOK     I.  27 


PROPOSITION  XVI.— THEOREM. 

60.  If  two  angles  have  their  sides  respectively  parallel  and  lying  in 
the  same  direction,  they  are  equal. 

Let  the  angles  ABC,  DEF,  have  their  sides  BA  and  ED  parallel 
and  in  the  same  direction,  and  also  their  sides 

BC  and  EF  parallel  and  in  the  same  direc-  ,          A          D 

tion.     Then  ABC  =  DEF.  \/    G/ 

For,  let  DE,  produced  if  necessary,  inter-  \       / 

sect  BC  in  G.     The  angle  DGC  is  equal  to      Ft -)f- F 

its  corresponding  angle  ABC  and  also  to  its  /  E\ 

corresponding   angle   DEF  (51) ;    therefore  j>, 

ABC  =  DEF. 

Note.  Two  parallels,  as  BA  and  ED,  are  said  to  be  in  the  same 
direction  when  they  lie  on  the  same  side  of  the  indefinite  straight 
line  joining  the  origins,  B  and  E,  of  these  parallels. 

61.  Corollary  I.    Two  angles,  as  ABC  and  D'EF',  having  their 
sides  parallel  and  lying  in  opposite  directions  (that  is  ED'  opposite 
to    BA    and    EF'   opposite   to   BC},   are   equal.      For   we   have 
D'EF'  =  DEF=  ABC. 

62.  Corollary  II.  Two  angles,  as  ABC  and  DEF',  having  two  of 
their  sides,  BA  and  ED,  parallel  and  in  the  same  direction,  while 
their  other  two  sides,  BC  and  EF',  are  parallel  and  in  opposite 
directions,  are  supplements  of  each  other. 

63.  Corollary  III.   If  two  angles,  ABC,  DEF,  have  their  sides  per 
pendicular  each  to  each,  that  is,  AB  to  ED  and 

BC  to   EF,    they   are  either  equal  or  supple-  / 

mentary.     For,  suppose  the  angle  DEF  to  be  B/F 

revolved  into  the  position  HEK,  by  revolving 
ED  and  EF  each  through  a  right  angle ;  that 
is,  ED  through  the  right  angle  DEH  and  EF 
through  the  right  angle  FEK.  Then  EH 
being  perpendicular  to  ED  is  parallel  to  AB,  and  EK  being  perpen 
dicular  to  EF  is  parallel  to  BC  (44)  ;  therefore  HEK,  or  DEF,  is 
either  equal  to  ABC  by  (60)  or  (61),  or  it  is  the  supplement  of 
ABC  by  (62). 


K 


28 


GEOMETRY. 


TRIANGLES. 


64.  Definitions.  A  plane  triangle  is  a  portion  of  a  plane  bounded 
by  three  intersecting  lines ;  as  AB  C.    The  sides  of  the 

triangle  are  the  portions  of  the  bounding  lines  in 
cluded  between  the  points  of  intersection ;  viz.',  AB, 
BC,  CA.  The  angles  of  the  triangle  are  the  angles 
formed  by  the  sides  with  each  other;  viz.,  CAB,  ABO,  BOA.  The 
three  angular  points,  A,  B,  C,  which  are  the  vertices  of  the  angles, 
are  also  called  the  vertices  of  the  triangle. 

If  a  side  of  a  triangle  is  produced,  the  angle  xA 

which   the  prolongation  makes  with  the  adjacent       /         \ 
side  is  called  an  exterior  angle;  as  A  CD.  B  °     D 

65.  A  triangle  is  called  scalene  (ABC)  when  no  two  of  its  sides 
are  equal ;  isosceles  (DEF)  when  two  of  its  sides  are  equal ;  equilat 
eral  (  GHI)  when  its  three  sides  are  equal. 


N 


A  right  triangle  is  one  which  has  a  right  angle ;  as  MNP,  which  is 
right-angled  at  N.  The  side  MP,  opposite  to  the  right  angle,  is  called 
the  hypotenuse. 

The  base  of  a  triangle  is  the  side  upon  which  it  is  supposed  to 
stand.  In  general  any  side  may  be  assumed  as  thd-base ;  but  in  an 
isosceles  triangle  DEF,  whose  sides  D^and  D^are  equal,  the  third 
side  EF  is  always  called  the  base. 

When  any  side  BC  of  a  triangle  has  been 
adopted  as  the  base,  the  angle  BA  C  opposite  to 
it  is  called  the  vertical  angle,  and  its  angular 
point  A  the  vertex  of  the  triangle.  The  per 
pendicular  AD  let  fall  from  the  vertex  upon  the  base  is  then  called 
the  altitude  of  the  triangle. 


B  O  O  K     1 .  29 


PKOPOSITION  XVII.— THEOKEM. 

66.  Any  side  of  a  triangle  is  less  than  the  sum  of  the  other  two. 
Let  BC  be  any  side  of  a  triangle  whose  other  two 

sides  are  AB  and  AC;  then  BC  <  AB  -f  AC. 
For,  the  straight  line  B  C  is  the  shortest  distance  be 
tween  the  points  B  and  C. 

67.  Corollary.  Any  side  of  a  triangle  is  greater  than  the  difference 
of  the  other  two.     For,  if  from  each  member  of  the  inequality 

BC<  AB  +  AC 
we  subtract  AB,  we  shall  have 

BC-AB<  AC,  or  AC>  BC—AB. 


PROPOSITION  XVIII.— THEOREM. 

68.  The  sum  of  the  three  angles  of  any  triangle  is  equal  to  two  right 
angles. 

Let  ABC  be  any  triangle ;  then,  the  sum  of  its  three  angles,  A,  B 
and  0,  is  equal  to  two  right  angles. 

For,  produce  BC  to  D,  and  draw  CE  par 
allel  to  BA.  The  angle  A  CE  is  equal  to  its 
alternate  angle  BAG  (49),  and  the  angle 
ECD  is  equal  to  its  corresponding  angle 

ABC  (51).     Therefore  the  sum  of  the  three  angles  of  the  triangle 
is  equal  to  ECD  -f-  ACE  -f  BCA,  which  is  two  right  angles  (14). 

69.  CorollaTy  I.  Any  exterior  angle,  as  A  CD,  is  equal  to  the  sum 
of  the  two  opposite  interior  angles,  A  and  B;  and   consequently 
greater  than  either  of  them. 

70.  Corollary  II.  If  one  angle  of  a  triangle  is  a  right  angle,  or  an 
obtuse  angle,  eaeh  of  the  other  two  angles  must  be  acute ;  that  is,  a 
triangle  cannot  have  two  right  angles,  or  two  obtuse  angles. 

71.  Corollary  III.  In  a  right  triangle,  the  sum  of  the  two  acute 
angles  is  equal  to  one  right  angle ;  that  is,  each  acute  angle  is  the 
complement  of  the  other  (18). 

72.  Corollary  IV.  If  two  angles  of  a  triangle  are  given,  or  only 
their  sum,  the  third  angle  will  be  found  by  subtracting  their  sum 
from  two  right  angles. 

3*  ' 


30  GEOMETRY. 

73.  Corollary  V.   If  two  angles  of  one  triangle  are  respectively 
equal  to  two  angles  of  another  triangle,  the  third  angle  of  the  one 
is  also  equal  to  the  third  angle  of  the  other. 

PKOPOSITION  XIX.— THEOEEM. 

74.  The  angle  contained  by  two  straight  lines  drawn  from  any  point 
within  a  triangle  to  the  extremities  of  one  of  the  sides  is  greater  than 
the  angle  contained  by  the  other  two  sides  of  the  triangle. 

From  any  point  D,  within  the  triangle  ABC,  let 
DB,  DC  be  drawn;  then,  the  angle  BDC  is  greater 
than  the  angle  BAG. 

For,  produce  BD  to  meet  A  C  in  E.     We  have  the 
angle  .BDO.REC' (69),  and  the  angle BEC> BAG]       B 
hence  BDC >  BAG. 

75.  Definition.  Equal  triangles,  and  in  general  equal  figures,  are 
those  which  can  exactly  fill  the  same  space,  or  which  can  be  applied 
to  each  other  so  as  to  coincide  in  all  their  parts. 

PROPOSITION  XX.— THEOREM. 

76.  Two  triangles  are  equal  when  two  sides  and  the  included  angle 
of  the  one  are  respectively  equal  to  two  sides  and  the  included  angle  of 
the  other. 

In  the  triangles  ABC,  DEF,  let  AB  be  equal  to  DE,  SO  to  EF, 
and  the  included  angle  B 

equal  to  the  included  angle     A  D  D 

\\  \\  *'* 

E;  then,  the  triangles  are 

equal. 

For,  the   triangle  ABC 
may    be   superposed    upon 

the  triangle  DEF,  by  applying  the  angle  B  to  the  equal  angle  E,  the 
side  BA  upon  its  equal  ED,  and  the  side  BC  upon  its  equal  EF. 
The  points  A  and  C  then  coinciding  with  the  points  D  and  F,  the 
side  A  C  will  coincide  with  the  side  DF,  and  the  triangles  will  coin 
cide  in  all  their  parts  ;  therefore  they  are  equal  (75). 

77.  Corollary.  If  in   two  triangles  ABC,  DEF,  there  are  given 
B=E,AB  =  DE  smdBC  =  EF,  there  will  follow'.!  =  D,  C  =  F, 


BOOK     I. 


31 


C    E 


F  F 


PROPOSITION  XXI.— THEOEEM. 

78.  Two  triangles  are  equal  when  a  side  and  the  two  adjacent  angles 
of  the  one  are  respectively  equal  to  a  side  and  the  two  adjacent  angles 
of  the  other. 

In  the  triangles  ABC,  DEF,  let  EC  be  equal  to  EF,  and  let  the 
angles  B  and  C  adjacent  to 
BC  be  respectively  equal  to 
the  angles  E  and  F  adja 
cent  to  EF-,  then,  the  tri 
angles  are  equal. 

For,  the  triangle  ABC 

may  be  superposed  upon  the  triangle  DEF,  by  applying  BC  to  its 
equal  EF,  the  point  B  upon  E,  and  the  point  C  upon  F.  The  angle 
B  being  equal  to  the  angle  E,  the  side  BA  will  take  the  direction  of 
ED,  and  the  point  A  will  fall  somewhere  in  the  line  ED.  The  angle 
C  being  equal  to  the  angle  F,  the  side  CA  will  take  the  direction  of 
FD,  and  the  point  A  will  fall  somewhere  in  the  line  FD.  Hence 
the  point  A,  falling  at  once  in  both  the  lines  ED  and  FD,  must  fall 
at  their  intersection  D.  Therefore  the  triangles  will  coincide  through 
out,  and  are  equal. 

79.  Corollary.  If  in  two  triangles  ABC,  DEF,  there  are  given 
B  ==  E,  C  =  F,  and  BC  =  EF,  there  will  follow  A  =  D,  AB  = 
DE, 


PROPOSITION   XXII.— THEOEEM. 

80.   Two  triangles  are  equal  when  the  three  sides  of  the  one  are  re 
spectively  equal  to  the  three  sides  of  the  other. 

In  the  triangles  ABC,  DEF,  let  AB  be  equal  to  DE,  AC  to  DF, 
and  BC  to  EF-,  then,  the  triangles 
are  equal. 

For,  suppose  the  triangle  ABC  to 
be  placed  so  that  its  base  BC  coin-    B 
cides  with  its  equal  EF,  but  so  that 
the  vertex  A  falls  on  the  opposite  side 
of  EF  from  D,  as  at  G;  and  join  DG  which  intersects  EF  in  H. 


32  GEOMETRY. 

Then,  by  hypothesis,  EG  =  ED  and  FG  =  ED ;  therefore,  E  and 
F  being  two  points  equally  distant  from  D  and  G,  the  line  EF  is 
perpendicular  to  DG  at  its  middle  point  H.     Hence,  if  the  figure 
DEF  be  revolved  upon  the  line  EF, 
H  being  a  fixed  point,  HD  will  fall 
upon  its  equal  H  G,  and  the  triangle 
DEF  will  coincide  entirely  with  the 
triangle  GEF.     Therefore,  the   tri 
angle  DEF  is  equal  to  the  triangle 
GEF,  or  to  the  triangle  ABC. 

81.  Corollary.  If  in  two  triangles  ABC,  DEF,  there  are  given 
AB  =  DE,AC=DF,BC  =  EF,  there  will  follow  A  =  D,B=E, 
C=F. 

82.  Scholium.  In  two  equal  triangles,  the  equal  angles  lie  opposite 
to  the  equal  sides. 


PROPOSITION  XXIII.— THEOREM. 

83.  Two  right  triangles  are  equal,  1st,  when  the  hypotenuse  and  a 
side  of  the  one  are  respectively  equal  to  the  hypotenuse  and  a  side  of  the 
other ;  or,  2d,  when  the  hypotenuse  and  an  acute  angle  of  the  one  are 
respectively  equal  to  the  hypotenuse  and  an  acute  angle  of  the  other. 

1st.  In  the  right  triangles  ABC, 
DEF,  let  the  hypotenuse  AB  be 
equal  to  DE,  and  the  side  A  C  to 
DF',  then,  the  triangles  are  equal. 

For,  applying  AC  to  its  equal 
DF,  the  angles  C  and  F  being 

equal,  the  side  CB  will  take  the  direction  FE,  and  B  will  fall  some 
where  in  the  line  FE.  But  AB  being  equal  to  DE,  will  cut  off  on 
FE  the  same  distance  from  the  perpendicular  (36),  and  hence  B  will 
fall  at  E.  The  triangles  will  therefore  coincide,  and  are  equal. 

2d.  Let  AB  =  DE,  and  the  angle  ABC  =  the  angle  DEF;  then, 
the  triangles  are  equal. 

For,  the  third  angles  BAG  and  EDF  are  equal  (73),  and  hence 
the  triangles  are  equal  by  (78). 


BOOK     I.  33 

* 

PROPOSITION  XXIV.— THEOREM. 

84.  If  two  sides  of  a  triangle  are  respectively  equal  to  two  sides  of 
another,  but  the  included  angle  in  the  first  triangle  is  greater  than  the 
included  angle  in  the  second,  the  third  side  of  the  first  triangle  is  greater 
than  the  third  side  of  the  second. 

Let  ABC  and  ABD  be  the  two  triangles  in  which  the  sides  AB, 
A  C  are  respectively  equal  to  the  sides  AB,  AD, 
but  the  included  angle  BAG  is  greater  than  the 
included  angle  BAD ;  then,  BC  is  greater  than 
BD. 

For,  suppose  the  line  AE  to  be  drawn,  bisect 
ing   the  angle  CAD  and  meeting  BC  in  E; 
join  DE.     The  triangles  AED,  AEG  are  equal  (76),  and  therefore 
ED  —  EC.     But  in  the  triangle  BDE  we  have 

BE  -}-  ED  >  BD, 
and  substituting  EC  for  its  equal  ED, 

BE  +  EC  >  BD,  or  BC  >  BD. 

85.  Corollary.  Conversely,  if  in   two  triangles  ABC,  DEF,  we 
have  AB  =  DE,  A  C  =  DF,  but  BC  >  EF;  then,  A  >  D. 

For,  if  A  were  equal  to  D,  we  should 
have  BC  =  EF  (76) ;  and  if  A  were  less 
than  D,  we  should  have  BC  <  EF  (by  the 
above  proposition)  ;  but  as  both  these  conclu 
sions  are  absurd,  being  contrary  to  the  hy 
pothesis,  we  can  only  have  A  >  D. 

PROPOSITION  XXV.— THEOREM. 

86.  In  an  isosceles  triangle,  the  angles  opposite  to  the  equal  sides  are 


Let  AB  and  AC  be  the  equal  sides  of  the  isosceles  triangle  ABC; 
then,  the  angles  B  and  C  are  equal.  A 

For,  let  D  be  the  middle  point  of  BC,  and  draw  AD. 
The  triangles  ABD  and  AD  C  are  equal  (80)  ;  therefore 
the  angle  ABD  =  the  angle  A  CD  (82). 

87.  Corollary  I.  From  the  equality  of  the  triangles 
ABD  and  A  CD,  we  also  have  the  angles  ADB  =  ADC, 

c 


34  GEOMETRY. 

and  BAD  =  CAD ;  that  is,  the  straight  line  joining  the  vertex  and 
the  middle  of  the  base  of  an  isosceles  triangle  is  perpendicular  to  the 
base  and  bisects  the  vertical  angle. 

Hence,  also,  the  straight  line  which  bisects  the  vertical  angle  of  an 
isosceles  triangle  bisects  the  base  at  right  angles. 

88.  Corollary  II.  Every  equilateral  triangle  is  also  equiangular ; 
and  by  (68),  each  of  its  angles  is  equal  to  one-third  of  two  right 
angles,  or  to  two-thirds  of  one  right  angle. 


PROPOSITION  XXVI.— THEOREM. 

89.  If  two  sides  of  a  triangle  are  unequal,  the  angles  opposite  to  them 
are  unequal,  and  the  greater  angle  is  opposite  to  the  greater  side. 

In  the  triangle  ABC,  let  AB  be  greater  than  AC', 
then,  the  angle  ACB  is  greater  than  the  angle  B. 

For,  from  the  greater  side  AB  cut  off  a  part  AD  — 
AC,  and  join  CD.     The  triangle  ADC  is  isosceles, 
and  therefore  the  angles  AD  C  and  ACD  are  equal 
(86).     But  the  whole  angle  ACB  is  greater  than  its 
part  ACD,  and  therefore  greater  than  ADC',  and  ADC,  an  exterior 
angle  of  the  triangle  BDC,  is  greater  than  the  angle  B  (69);  still 
more,  then,  is  ACB  greater  than  B. 


PROPOSITION  XXVII.— THEOREM. 

90.  If  two  angles  of  a  triangle  are  equal,  the  sides  opposite  to  them 
are  equal. 

In  the  triangle  ABC,  let  the  angles  B  and  C  be 
equal ;  then,  the  sides  AB  and  AC  are  equal. 

For,  if  the  sides  AB  and  AC  were  unequal,  the 
angles  B  and  C  could  not  be  equal  (89). 

91.  Corollary.  Every  equiangular  triangle  is  also  equilateral. 


BOOK    I.  35 


PKOPOSITION  XXVIII.— THEOKEM. 

92.  If  two  angles  of  a  triangle  are  unequal,  the  sides  opposite  to  them 
are  unequal,  and  the  greater  side  is  opposite  to  the  greater  angle. 

In  the  triangle  ABC  let  the  angle  C  be  greater  than 
the  angle  B ;  then,  AB  is  greater  than  A  C. 

For,  suppose  the  line  CD  to  be  drawn,  cutting  off 
from  the  greater  angle  a  part  BCD  =  B.  Then  BDC 
is  an  isosceles  triangle,  and  D  C  =  DB.  But  in  the 
triangle  ADC,  we  have  AD  -j-  DC>  AC',  or,  putting 
DB  for  its  equal  DC,  AD  +  DB  >  AC',  or  AB  >  A  C. 


POLYGONS. 

93.  Definitions.  A  polygon  is  a  portion  of  a  plane  bounded  by 
straight  lines;  as  ABODE.  The  bounding  lines 
are  the  sides;  their  sum  is  the  perimeter  of  the 
polygon.  The  angles  which  the  adjacent  sides  make 
with  each  other  are  the  angles  of  the  polygon ;  and 
the  vertices  of  these  angles  are  called  the  vertices 
of  the  polygon. 

Any  line  joining  two  vertices  not  consecutive  is  called  a  diagonal; 
as  AC. 

94.  Definitions.  Polygons  are  classed  according  to  the  number  of 
their  sides : 

A  triangle  is  a  polygon  of  three  sides. 

A  quadrilateral  is  a  polygon  of  four  sides. 

A  pentagon  has  five  sides ;  a  hexagon,  six ;  a  heptagon,  seven  ;  an 
octagon,  eight ;  an  enneagon,  nine ;  a  decagon,  ten ;  a  dodecagon, 
twelve ;  etc. 

An  equilateral  polygon  is  one  all  of  whose  sides  are  equal;  an 
equiangular  polygon,  one  all  of  whose  angles  are  equal. 

95.  Definition.  A  convex  polygon  is  one  no  side  of  which  when 
produced  can  enter  within  the  space  enclosed  by  the  perimeter,  as 
ABCDE  in  (93).     Each  of  the  angles  of  such  a  polygon  is  less  than 
two  right  angles. 

It  is  also  evident  from  the  definition  that  the  perimeter  of  a  convex 


36 


GEOMETRY. 


polygon  cannot  be  intersected  by  a  straight  line  in  more  than  two 
points. 

A  concave  polygon  is  one  of  which  two  or 
more  sides,  when  produced,  will  enter  the  space 
enclosed  by  the  perimeter;  as  MNOPQ,  of 
which  OP  and  QP  when  produced  will  enter 
within  the  polygon.  The  angle  OPQ,  formed 
by  two  adjacent  re-entrant  sides,  is  called  a  re 
entrant  angle ;  and  hence  a  concave  polygon  is  sometimes  called  a 
re-entrant  polygon. 

All  the  polygons  hereafter  considered  will  be  understood  to  be 
convex. 

96.  A  polygon  may  be  divided  into  triangles  by  diagonals  drawn 
from    one    of   its    vertices.      Thus    the    pentagon 

ABODE  is  divided  into  three  triangles  by  the 
diagonals  drawn  from  A.  The  number  of  triangles 
into  which  any  polygon  can  thus  be  divided  is  evi 
dently  equal  to  the  number  of  its  sides,  less  two. 
The  number  of  diagonals  so  drawn  is  equal  to  the 
number  of  sides,  less  three. 

97.  Two     polygons     ABODE, 
A'B'C'D'E',  are  equal  when  they 
can  be  divided  by  diagonals  into  the 
same  number  of  triangles,  equal  each 
to  each,  and  similarly  arranged ;  for 
the  polygons  can  evidently  be  super 
posed,  one  upon  the  other,  so  as  to  coincide. 

98.  Definitions.  Two  polygons 
are   mutually  equiangular  when 
the  angles   of   the  one  are   re 
spectively  equal   to   the   angles 
of  the  other,  taken  in  the  same 
order;  as  ABCD,  A'B'C'D',  in 
which  A  =  A',  B  =  B',  etc. 

The  equal  angles  are  called  homologous  angles;  the  sides  containing 
equal  angles,  and  similarly  placed,  are  homologous  sides;  thus 
A  and  A'  are  homologous  angles,  AB  and  A'B'  are  homologous 
sides,  etc. 


D> 


B' 


B  O  O  K     1 .  37 

Two  polygons  are  mutually  equilateral  when  the  sides  of  the  one 
are  respectively  equal  to  the  sides  of 
the  other,  taken  in  the  same  order ; 
as    MNPQ,    M'N'P'Q',   in    which 
MN  =  M'N',  NP  =  N'P',  etc. 
The  equal  sides  are  homologous ;  and 
angles  contained  by  equal  sides  simi 
larly  placed,  are  homologous ;  thus  MN  and  M'N'  are  homologous 
sides ;  M  and  M'  are  homologous  angles. 

Two  mutually  equiangular  polygons  are  not  necessarily  also  mu 
tually  equilateral.  Nor  are  two  mutually  equilateral  polygons 
necessarily  also  mutually  equiangular,  except  in  the  case  of  tri 
angles  (80). 

If  two  polygons  are  mutually  equilateral  and  also  mutually  equi 
angular,  they  are  equal ;  for  they  can  evidently  be  superposed,  one 
upon  the  other,  so  as  to  coincide. 


PKOPOSITION  XXIX.— THEOREM. 

99.  The  sum  of  all  the  angles  of  any  polygon  is  equal  to  two  right 
angles  taken  as  many  times  less  two  as  the  polygon  has  sides. 

For,  by  drawing  diagonals  from  any  one  vertex,  the  polygon  can 
be  divided  into  as  many  triangles  as  it  has  sides,  less  two  (96).  The 
sum  of  the  angles  of  all  the  triangles  is  the  same  as  the  sum  of  the 
angles  of  the  polygon,  and  the  sum  of  the  angles  of  each  triangle  is 
two  right  angles  (68).  Therefore,  the  sum  of  the  angles  of  the 
polygon  is  two  right  angles  taken  as  many  times  less  two  as  the 
polygon  has  sides. 

100.  Corollary  I.    If  N  denotes  the  number  of  the  sides  of  the 
polygon,   and  E  a  right  angle,  the  sum  of  the  angles  is   2E  X 
(N—  2)  ==  (2N—  ±)R=  2NE  —  4R;  that  is,  twice  as  many 
right  angles  as  the  polygon  has  sides,  less  four  right  angles. 

For  example,  the  sum  of  the  angles  of  a  quadrilateral  is  fou- 
right  angles ;  of  a  pentagon,  six  right  angles ;  of  a  hexagon,  eight 
right  angles,  etc. 
4 


38  G  E  O  M  E  T  R  Y. 

101.   Corollary  II.    If  all  the  sides  of  any  polygon  ABODE,  be 
produced  so  as  to  form  one  exterior  angle  at 
each  vertex,  the  sum  of  these  exterior  angles,  d] 

a,  b,  G,  d,  e,  is  four  right  angles.     For,  the  sum 
of  each  interior  and  its  adjacent  exterior  angle, 
as  A  -f-  a,  is  two  right  angles  (11) ;  therefore, 
the  sum  of  all   the  angles,   both  interior  and 
exterior,  is  twice  as  many  right  angles  as  the 
polygon  has  sides.     But  the  sum  of  the  interior  angles  alone  is  twice 
as  many  right  angles  as  the  polygon  has  sides,  less  four  right  angles 
(100)  ;  therefore  the  sum  of  the  exterior  angles  is  equal  to  four  right 
angles. 

This  is  also  proved  in  a  very  simple  manner,  by  drawing,  from 
any  point  in  the  plane  of  the  polygon,  a  series  of  lines  respectively 
parallel  to  the  sides  of  the  polygon  and  in  the  same  directions  as 
their  prolongations.  The  consecutive  angles  formed  by  these  lines 
will  be  equal  to  the  exterior  angles  of  the  polygon  (60),  and  their 
sum  is  four  right  angles  (15). 


QUADRILATERALS. 

102.  Definitions.  Quadrilaterals  are  divided  into  classes  as  follows : 
1st.  The  trapezium  (A)  which  has  no  two  of  its 


sides  parallel. 


2d.  The  trapezoid  (E)  which  has  two  sides  par 
allel.  The  parallel  sides  are  called  the  bases,  and 
the  perpendicular  distance  between  them  the  alti 
tude  of  the  trapezoid. 

3d.  The  parallelogram  ((7)  which  is  bounded  by      r r~v 

two  pairs  of  parallel  sides.  \ \ 

The  side  upon  which  a  parallelogram  is  supposed 
to  stand  and  the  opposite  side  are  called  its  lower  and  upper  bases. 
The  perpendicular  distance  between  the  bases  is  the  altitude. 

103.    Definitions.    Parallelograms    are   divided    into    species,   as 
follows : 


B  O  O  K      1 .  39 


1st.  The  rhomboid  (a),  whose   adjacent  sides       ^ 
are  not  equal  and  whose  angles  are  not  right 
angles. 

2d.  The  rhombus,  or  lozenge  (6),  whose  sides  are 
all  equal. 

3d.  The  rectangle  (c),  whose  angles  are  all  equal 
and  therefore  right  angles. 

4th.  The  square  (c?),  whose  sides  are  all  equal  and  whose 
angles  are  all  equal. 

The  square  is  at  once  a  rhombus  and  a  rectangle. 


PEOPOSITION  XXX.— THEOEEM. 

104.  In  every  parallelogram,  the  opposite  angles  are  equal,  and  the 
opposite  sides  are  equal. 

Let  ABCD  be  a  parallelogram. 

1st.  The  opposite  angles  B  and  D,  contained 

by  parallel  lines  lying  in  opposite  directions,  /<r -7 

are  equal  (61) ;   and  for  the  same  reason  the          /      \      / 
opposite  angles  A  and  C  are  equal.  B  '<? 

2d.  Draw  the  diagonal  AC.     Since  AD  and 

J5(7are  parallel,  the  alternate  angles  CAD  and  ACB  are  equal  (49), 
and  since  DO  and  AB  are  parallel,  the  alternate  angles  A  CD  and 
CAB  are  equal.  Therefore,  the  triangles  ADC  and  CBA  are  equal 
(78),  and  the  sides  opposite  to  the  equal  angles  are  equal,  namely, 
AD  =  BC,  and  DC  =  AB. 

105.  Corollary  I.  A  diagonal  of  a  parallelogram  divides  it  into 
two  equal  triangles. 

106.  Corollary  II.    If  one  angle  of  a   parallelogram   is  a  right 
angle,  all  its  angles  are  right  angles,  and  the  figure  is  a  rectangle. 


40  GEOMETRY. 


PROPOSITION  XXXI.—  THEOREM. 

107.  If  the  opposite  angles  of  a  quadrilateral  are  equal,  or  if  its 
opposite  sides  are  equal,  the  figure  is  a  parallelogram. 

1st.  Let  the  opposite  angles  of  the  quadrilateral  ABCD  be  equal, 
or  A  =  C  and  B  =  D.     Then,  by  adding  equals, 
we  have  "  -  \ 


therefore,  each  of  the  sums  A  -f  B  and  C  -\-  D 
is  equal  to  one-half  the  sum  of  the  four  angles.  But  the  sum  of  the 
four  angles  is  equal  to  four  right  angles  (100)  ;  therefore,  A  -f  B  is 
equal  to  two  right  angles,  and  the  lines  AD  and  B  C  are  parallel  (56). 
In  like  manner  it  may  be  proved  that  AB  and  CD  are  parallel. 
Therefore  the  figure  is  a  parallelogram. 

2d.  Let  the  opposite  sides  of  the  quadrilateral  ABCD  be  equal, 
or  BC  =  AD  and  AB  =  DC.     Then,  drawing 
the  diagonal  AC,  the  triangles  ABC,  A  CD  are 
equal  (80)  ;  therefore,  the  angles  CAD  and  ACB 
are  equal,  and  the  lines  AD  and  BC  are  parallel 
(54).     Also  since  the  angles  CAB  and  A  CD  are 
equal,  the  lines  AB  and  DC  are  parallel.     Therefore  ABCD  is  a 
parallelogram. 

PROPOSITION  XXXII.—  THEOREM. 

108.  If  two  opposite  sides  of  a  quadrilateral  are  equal  and  parallel, 
the  figure  is  a  parallelogram. 

Let   the  opposite  sides  BC  and  AD  of  the 

quadrilateral   ABCD   be    equal   and    parallel.  /--  -  y 

Draw  the  diagonal  AC.     The  alternate  angles          /      \y 
CAD  and  A  CB  are  equal  (49),  and  hence  the      B  c 

triangles  ADC  and  CBA  are  equal  (76).    There 
fore,  the  sides  AB  and  CD  are  equal  and  the  figure  is  a  parallelo 
gram  (107). 


BOOK     I.  41 


PKOPOSITION  XXXIII.— THEOREM. 

109.  The  diagonals  of  a  parallelogram  bisect  each  other. 

Let  the  diagonals  A  C,  BD  of  the  parallelogram  AB  CD  intersect 
in  E]  then,  AE  =  EC  and  ED  =  EB. 

For,  the  side  AD  and  the  angles  EAD,  ADE, 
of  the  triangle  EAD,  are  respectively  equal  to 

the  side  CB  and  the  angles  ECB,  EBC  of  the       £*- ^/ 

triangle  ECB',  hence  these  triangles  are  equal 

(78),  and  the  sides  respectively  opposite  the  equal  angles  are  equal, 

namely,  AE  =  EC  and  ED  =  EB. 

110.  Corollary  I.  The  diagonals  of  a  rhombus  ABCD  bisect  each 
other  at  right  angles  in  E.     For,  since  AD  =  CD 

and  AE  —  EC,  ED  is  perpendicular  to  AC  (41). 

111.  Corollary  II.    The  diagonals  of  a  rhombus 
bisect  its  opposite  angles.     For,  in  each  of  the  isos 
celes  triangles  ADC,  ABC,  BCD,  DAB,  the  line 
drawn  from  the  vertex  to  the  middle  of  the  base 
bisects  the  vertical  angle  (87). 


PROPOSITION  XXXIV.— THEOREM. 

112.  If  the  diagonals  of  a  quadrilateral  bisect  each  other,  the  figure 
is  a  parallelogram. 

Let  the  diagonals  of  the  quadrilateral  ABCD  bisect  each  other 
in  E.     Then,  the  triangles  AED  and  CEB  are 
equal  (76),  and  the  angles  EAD,  ECB,  respect-  /x  E  ^.,--1 

ively  opposite  the  equal  sides,  are  equal.    There-          /£-•**      \  / 
fore  AD  and  BC  are  parallel   (54).     In  like         B"  V 

manner  AB  and  DC  are  shown  to  be  parallel, 
and  the  figure  is  a  parallelogram. 

113.  Corollary.    If  the  diagonals  of  a  quadrilateral  bisect  each 
other  at  right  angles,  the  figure  is  a  rhombus. 


42  GEOMETRY. 


PKOPOSITION   XXXV.— THEOKEM. 

114.   The  diagonals  of  a  rectangle  are  equal 

Let  ABCD  be  a  rectangle;  then  its  diagonals,  AC  and  BD,  are 
equal. 

-A 

For,  the  right  triangles  ABC  and  DCB  are  equal 
(76)  ;  therefore,  AC  =  BD. 


115.  Corollary  I.  The  diagonals  of  a  square  are  equal, 
and,  since  the  square  is  also  a  rhombus,  they  bisect  each 
other  at  right  angles  (110),  and  also  bisect  the  angles 
of  the  square  (111). 

116.  Corollary  II.  A  parallelogram  is  a  rectangle  if  its  diagonals 
are  equal. 

117.  Corollary  III.  A  quadrilateral  is  a  square,  if  its  diagonals 
are  equal  and  bisect  each  other  at  right  angles. 

118.  Scholium.  The  rectangle,  being  a  species  of  parallelogram, 
has  all  the  properties  of  a  parallelogram. 

The  square,  being  at  once  a  parallelogram,  a  rectangle  and  a 
rhombus,  has  the  properties  of  all  these  figures. 


PKOPOSITION  XXXVI.— THEOKEM. 

119.  Two  parallelograms  are  equal  when  two  adjacent  sides  and  the 
included  angle,  of  the  one  are  equal  to  two  adjacent  sides  and  the 
included  angle  of  the  other. 

Let  AC,  AC',  have  AB  =  A'B', »      D *    D' c' 

AD  =  AD',  and  the  angle  BAD  =       I /    / / 

B'A'D';    then,   these   parallelograms  A                B    A>                B< 
are  equal. 

For  they  may  evidently  be  applied  the  one  to  the  other  so  as  to 
coincide  throughout. 

120.  Corollary.    Two   rectangles    are 
equal  when  they  have  equal  bases  and 
equal  altitudes. 


BOOK     I.  43 

APPLICATIONS. 

PKOPOSITION  XXXVII.— THEOEEM. 

121.  If  a  straight  line  drawn  parallel  to  the  base  of  a  triangle  bisects 
one  of  the  sides,  it  also  bisects  the  other  side;  and  the  portion  of  it 
intercepted  between  the  two  sides  is  equal  to  one-half  the  base. 

Let  DE  be  parallel  to  the  base  -BO  of  the  triangle 
ABC,  and  bisect  the  side  AB  in  D;  then,  it  bisects 
the  side  AC  in  E,  and  DE  =  iB<7. 

1st.  Through  D  suppose  DF  to  be  drawn  parallel 
to  AC.  In  the  triangles  ADE,  DBF,  we  have 
AD  —  DB,  and  the  angles  adjacent  to  these  sides 
equal,  namely  DAE  =  BDF,  and  ADE  =  DBF  (51) ;  therefore 
these  triangles  are  equal  (78),  and  AE  =  DF.  Also,  since  DECF 
is  a  parallelogram,  DF  =  EC  (104)  ;  and  hence  AE  =  EC. 

2d.  The  triangles  ADE  and  BDF  being  equal,  we  have  DE  =  BF, 
and  in  the  parallelogram  DECF  we  have  DE  —  EC;  therefore 
BF  =  FC.  Hence  F  is  the  middle  point  of  BC,andDE  =  \E  C. 

122.  Corollary  I.   The  straight  line  DE,  joining  the  middle  points 
of  the  sides  AB,  A  C,  of  the  triangle  AB  C,  is  parallel  to  third  side  B  C, 
and  is  equal  to  one-half  of  BC.     For,  the  straight  line  drawn  through 
D  parallel  to  BC,  passes  through  E  (121),  and  is  therefore  identical 
with  DE.     Consequently,  also,  DE  =  %BC. 

123.  Corollary  II.    The  straight  line  drawn  parallel  to  the  bases  of  a 
trapezoid,  bisecting  one  of  the  non-parallel  sides,  also  bisects  the  opposite 
side. 

Let  AB  CD  be  a  trapezoid,  BC  and  AD  its  A^ 

parallel  bases,  E  the  middle  point  of  AB,  and  ^— -7x7 

let  EF  be  drawn  parallel  to  BCor  AD ;  then,        /  " 

.Fis  the  middle  of  DC.     For,  draw  the  diago 
nal  AC,  intersecting  EF  hi  H.     Then  in  the  triangle  ABC,  EH  is 
drawn  through  the  middle  of  AB  parallel  to  BC;  therefore  H  is 
the  middle  of  AC.     In  the  triangle  ACD,  HF  is  drawn  through 
the  middle  of  AC  parallel  to  AD-,  therefore  F  is  the  middle  of  DC. 

124.  Corollary  III.    In  a  trapezoid,  the  straight  line  joining  the 
middle  points  of  the  non-parallel  sides  is  parallel  to  the  bases,  and  is 
equal  to  one-half  their  sum. 


44 


GEOMETRY. 


Let  EF  join  the  middle  points,  E  and  F, 
of  AB  and  DC.     Then,  1st,  EF  is  parallel 
to  BC.     For,   by  Cor.  II.  the  straight  line          / 
drawn    through    E   parallel    to   BC   passes      $  — 
through  F  and  is  therefore  identical  with  EF. 

2d.  Drawing  the  diagonal  AC,  intersecting  EF  in  H,  we  have,  in 
the  triangle  ABC, 

and  in  the  triangle  ACD, 


the  sum  of  which  gives 

EF= 


PKOPOSITION  XXXVIII.— THEOKEM. 

125.  If  a  series  of  parallels  cutting  any  two  straight  lines  intercept 
equal  distances  on  one  of  these  lines,  they  also  intercept  equal  distances 
on  the  other  line. 

Let  MNj  M'N',  be  two  straight  lines  cut  by  a  series  of  parallels 
AA',  BB',  CC',  DD'-,  then,  if  AB,  BC,  CD  are  equal,  A'B',  B'C', 
C'D'  are  also  equal. 

For,  through  the  points  A,  B,  C,  draw  Ab, 
Be,  Cd,  parallel  to  M'N'.  In  the  triangles 
ABb,  BCc,  CDd,  we  have  AB  =  BC  =  CD-, 
and  the  corresponding  angles  adjacent  to  these 
sides  are  equal  (51),  namely,  BAb  =  CBc  = 
DCd, and  ABb  =  BCc  =  CDd',  therefore, these 
triangles  are  equal  to  each  other  (78),  and  Ab 

-  Be  =  Cd.  But,  the  figures  A'b,  B'c,  C'd,  being  parallelo 
grams,  we  have  Ab  =  A'B',  Be  =  B'C',  Cd  =  C'D';  therefore, 
AB'  =  B'C'  =  C'D'. 


PKOPOSITION  XXXIX.— THEOKEM. 

126.  Every  point  in  the  bisector  of  an  angle  is  equally  distant  from 
the  sides  of  the  angle ;  and  every  point  not  in  the  bisector,  but  within  the 
angle,  is  unequally  distant  from  the  sides  of  the  angle. 


BOOK     I. 


45 


1st.  Let  AD  be  the  bisector  of  the  angle 
BA  C,  P  any  point  in  it,  and  PE,  PF,  the  per 
pendicular  distances  of  Pfrom  AB  and  AC; 
then,  PE  =  PF. 

For,  the  right  triangles  APE,  APF,  having 
the  angles  PAE  and  PAF  equal,  and  AP  com 
mon,  are  equal  (83)  ;  therefore,  PE  =  PF. 

2d.  Let  Q  be  any  point  not  in  the  bisector,  but  within  the  angle ; 
then,  the  perpendicular  distances  QE  and  QH  are  unequal. 

For,  suppose  that  one  of  these  distances,  as  QE,  cuts  the  bisector 
in  some  point  P:  from  P  let  PF  be  drawn  perpendicular  to  AC, 
and  join  QF.  We  have  QH  <  QF;  also  QF  <  QP  +  PF,  or 
QF<QP-\-  PE,  or  QF  <  QE;  therefore,  QH  <  QE. 

When  the  angle  BAG  is  obtuse,  the 
point  Q,  not  in  the  bisector,  may  be  so 
situated  that  the  perpendicular  on  one  of 
the  sides,  as  AB,  will  fall  at  the  vertex  A ; 
the  perpendicular  QH  is  then  less  than 
the  oblique  line  QA.  Or,  a  point  Q'  may 

be  so  situated  that  the  perpendicular  Q'E',  let  fall  on  one  of  the  sides, 
as  AB,  will  meet  that  side  produced  through  the  vertex  A;  this 
perpendicular  must  cut  the  side  A  C  in  some  point,  K,  and  we  then 
have  Q'H'  <  Q'K  <  Q'E'. 

127.  Corollary.  The  bisector  of  an  angle  is  the  locus  (40)  of  all 
the  points  within  the  angle  which  are  equally  distant  from  its  sides. 


PKOPOSITION  XL— THEOEEM. 

128.   The  three  bisectors  of  the  three  angles  of  a  triangle  meet  in  the 
same  point. 

Let  AD,  BE,  CF,  be  the  bisectors  of  the 
angles  A,  B,  C,  respectively,  of  the  triangle 
ABC. 

Let  the  two  bisectors  AD,  BE,  meet  in  0. 
The  point  0,  being  in  AD,  is  equally  dis 
tant  from  AB  and  AC  (126);  and  being 
in  BE,  it  is  equally  distant  from  AB  and  BC; 


46 


GEOMETRY. 


therefore,  the  point  0  is  equally  distant  from 
J.(7and  BC,  and  must  lie  in  the  bisector  of 
the  angle  C  (127).  That  is,  the  bisector  CF 
also  passes  through  0,  and  the  three  bisect 
ors  meet  in  the  same  point. 


129.  Corollary.  The  point  in  which  the  three  bisectors  of  the 
angles  of  a  triangle  meet  is  equally  distant  from  the  three  sides  of 
the  triangle. 


PROPOSITION  XLL— THEOREM. 

130.  The  three  perpendiculars  erected  at  the  middle  points  of  the 
sides  of  a  triangle  meet  in  the  same  point. 

Let  DG,  EH,  FK,  be  the  perpendiculars 
erected  to  BC,  CA,  AB,  respectively,  at  their 
middle  points,  D,  E,  F. 

It  is  first  necessary  to  prove  that  any  two  of 
these  perpendiculars,  as  DO,  EH,  meet  in  some 
point.     If  they  did  not  meet,  they  would  be 
parallel,  and  then  CB  and  CA  being  perpen 
diculars  to  these  parallels  from  the  same  point  C,  would  be  in  one 
straight  line,  which  is  impossible,  since  they  are  two  sides  of  a  tri 
angle.     Therefore,  DG  and  EH  are  not  parallel,  and  must  meet  in 
some  point,  as  0. 

Now  the  point  0  being  in  the  perpendicular  DGfe  equally  distant 
from  B  and  C  (38),  and  being  also  in  the  perpendicular  EH,  it  is 
equally  distant  from  A  and  C;  therefore  it  is  equally  distant  from  A 
and  B,  and  must  lie  in  the  perpendicular  FK  (39).  That  is,  the 
perpendicular  FK  passes  through  0,  and  the  three  perpendiculars 
meet  in  the  same  point. 

131.  Corollary.  The  point  in  which  the  three  perpendiculars  meet 
is  equally  distant  from  the  three  vertices  of  the  triangle. 


BOOK     I. 


47 


PKOPOSITION  XLII.— THEOREM. 

132.  The  three  perpendiculars  from  the  vertices  of  a  triangle  to  the 
opposite  sides  meet  in  the  same  point 

Let  AD,  BE,   CF,  be  the  perpen-      £ „.    ^  *' 

diculars  from  the  vertices  of  the  tri 
angle  ABC  to  the  opposite  sides,  re 
spectively. 

Through  the  three  vertices,  A,  B,  C, 
draw  the  lines  B'C',  AB',  AC',  re 
spectively  parallel   to  BC,  AB,  AC.  A' 
Then  the  two  quadrilaterals  ABCB' 

and  ACBC'  are  parallelograms,  and  we  have  AB'  =  BC  and 
AC'  =  BC',  therefore  AB'  =  AC',  or  A  is  the  middle  of  B'  C'. 
But  AD  being  perpendicular  to  BC  is  perpendicular  to  the  parallel 
B'C'',  therefore  AD  is  the  perpendicular  to  B' C'  erected  at  its 
middle  point  A.  In  like  manner,  it  is  shown  that  BE  and  CF  are 
the  perpendiculars  to  A  C'  and  AB'  at  their  middle  points ;  there 
fore,  by  (130),  the  three  perpendiculars  meet  in  the  same  point. 

133.  Definition.  A  straight  line  drawn  from  any  vertex  of  a  tri 
angle  to  the  middle  point  of  the  opposite  side  is 

called  a  medial  line  of  the  triangle.  Thus,  D  being 
the  middle  point  of  BC,  AD  is  the  medial  line  to 
BC.  if 


PROPOSITION  XLIIL— THEOREM. 

134.   The  three  medial  lines  of  a  triangle  meet  in  the  same  point. 

Let  D,  E,  F,  be  the  three  middle  points  of 
the  sides  of  the  triangle  ABC',  AD,  BE,  CF, 
the  three  medial  lines. 

Let  the  two  medial  lines,  AD  and  BE,  meet 
in  0.     Let  O  be  the  middle  point  of  OA,  and 
H  the  middle  point  of  OB-,  join  GH,  HD, 
DE,  EG.     In  the  triangle  A  OB,  GH  is  par 
allel  to  AB,  and  GH  =  %AB:    and  in  the  triangle  ABC,  ED  is 
parallel  to  AB,  and  ED  =  ±AB  (122).     Therefore,  HG  and  ED, 
being  parallel  to  AB,  are  parallel  to  each  other ;   and  each  being 


48 


GEOMETRY. 


equal  to  £ AB,  they  are  equal  to  each  other ;  consequently,  EGHD 

is   a   parallelogram   (108),  and    its   diagonals 

bisect  each  other  (109).     Therefore  OD  =  OG 

—  GA,  or  OD  =  ^AD ;  that  is,  the  medial 

line  BE  cuts  the  medial  line  AD  at  a  point  0 

whose  distance  from  D  is  one-third  of  AD.     In 

the  same  way  it  is  proved  that  the  medial  line 

CF  cuts  AD  at  a  point  whose  distance  from  D 

is  one-third  of  AD,  that  is,  at  the  same  point  0 ;  and  therefore  the 

three  medial  lines  meet  in  the  same  point. 


SYMMETRICAL  FIGURES. 
a.  Symmetry  with  respect  to  an  axis. 

135.  Definition.  Two  points  are  symmetrical  with  respect  to  a  fixed 
straight  line,  called  the  axis  of  symmetry,  when  this  axis  bisects  at 
right  angles  the  straight  line  joining  the  two  points. 

Thus,  A  and  A'  are  symmetrical  with  respect  to 
the  axis  MN,  if  MN  bisects  A  A'  at  right  angles 
at  a.  M 


If  the  portion  of  the  plane  containing  the  point 
A  on  one  side  of  the  axis  MN,  is  revolved  about 
this  axis  (or  folded  over)  until  it  coincides  with  the 
portion  on  the  other  side  of  the  axis,  the  point  A'  at  which  A  falls 
is  the  symmetrical  point  of  A. 

136.  Definition.  Any  two  figures  are  symmetrical  with  respect  to 
an  axis  when  every  point  of  one  figure  has  its  symmetrical  point  on 
the  other. 

Thus,  A'B'  is  the  symmetrical  figure 
of  the  straight  line  AB,  with  respect  to 
the  axis  MN,  every  point,  as  C,  of  the  one, 
having  its  symmetrical  point  C'  in  the 
other. 

The  symmetrical  figure  of  an  indefinite 
straight  line,  AB,  is  an  indefinite  straight 
line,  A'B',  which  intersects  the  first  in  the 
axis  and  makes  the  same  angle  with  the 
axis  as  the  first  line. 


BOOK     I. 


49 


137.  Definition.  In   two   symmetrical   figures   the   corresponding 
symmetrical  lines  are  called  homologous. 

Thus,  in  the  symmetrical  figures  ABODE, 
A'B'C'D'E',  the  homologous  lines  are  AB 
and  A'B',  BC and  B'C't  etc. 

In  all  cases,  two  figures,  symmetrical  with 
respect  to  an  axis,  can  be  brought  into  coin 
cidence  by  the  revolution  of  either  about  the 
axis. 

6.  Symmetry  with  respect  to  a  centre. 

138.  Definition.  Two  points  are  symmetrical  with  respect  to  a  fixed 
point,  called  the  centre  of  symmetry,  when   this  centre  bisects  the 
straight  line  joining  the  two  points. 

Thus,  A  and  A'  are  symmetrical  with  respect  ...-^ 

to  the  centre  0,  if  the  line  AA  passes  through  „••••'' 

0  and  is  bisected  at  0.  X' 

The  distance  of  a  point  from  the  centre  is  called        1 
its  radius  of  symmetry.     A  point  A  is  brought  into 
coincidence  with  its  symmetrical  point  A,  by  revolving  its  radius 
OA  through  two  right  angles  in  its  own  plane  (16). 

139.  Definition.  Any  two  figures  are  symmetrical  with  respect  to 
a  centre,  when  every  point  of  one  figure  has  its  symmetrical  point 
on  the  other. 

Thus,  A'B'  is  the  symmetrical 
figure  of  the  straight  line  AB  with 
respect  to  the  centre  0. 

Since  the  triangles  AOB,  A  OB', 
are  equal  (76),  the  angle  B  is  equal 

to  the  angle  B' ;  therefore,  AB  and  A'B'  are  parallel.  In  general, 
the  homologous  lines  of  two  figures, 
symmetrical  with  respect  to  a  centre, 
are  parallel.  Thus,  in  the  symmetri 
cal  figures  ABCD,  AB'C'D',  the 
homologous  lines  AB  and  A'B'  are 
parallel,  BC  and  B' C'  are  parallel, 
etc. 

Two  figures  symmetrical  with  respect  to  a  centre  can  be  brought 

5  D 


50 


GEOMETRY. 


into  coincidence  by  revolving  one  of  them,  in  its  own  plane,  about 
the  centre ;  every  radius  of  symmetry  revolving  through  two  right 
angles  at  the  same  time. 

140.  Definition.  Any  single  figure  is  called  a  symmetrical  figure, 
either  when  it  can  be  divided  by  an  axis  into  two  figures  symmetri 
cal  with  respect  to  that  axis,  or  when  it  has  a  centre  such  that  every 
straight  line  drawn  through  it  cuts  the  figure  in  two  points  which 
are  symmetrical  with  respect  to  this  centre. 

Thus,  ABCDC'B'  is  a  symmetrical 
figure  with  respect  to  the  axis  MN, 
being  divided  by  MN  into  two  figures, 
AB CD  and  AB'C'D,  which  are  sym 
metrical  with  respect  to  MN. 

Also,  the  figure  ABCDEF  is  symmetrical  with  respect  to  the 
centre  0,  its  vertices,  taken  two  and  two, 
being  symmetrical  with  respect  to  0.  In 
this  case,  any  straight  line  KL  drawn 
through  the  centre  and  terminated  by  the 
perimeter,  is  called  a  diameter. 


PROPOSITION  XLIV.— THEOREM. 

141.  If  a  figure  is  symmetrical  with  respect  to  two  AXES  perpendicular 
to  each  other,  it  is  also  symmetrical  with  respect  to  the  intersection  of 
these  axes  as  a  CENTRE  of  symmetry. 

Let  the  figure  ABCDEF GH  be 
symmetrical  with  respect  to  the  two 
perpendicular  axes  MN,  PQ,  which 
intersect  in  0 ;  then,  the  point  0  is 
also  the  centre  of  symmetry  of  the 
figure. 

For,  let  T  be  any  point  in  the 
perimeter  of  the  figure;  draw  TR T' 
perpendicular  to  MN,  and  TSt  per 
pendicular  to  PQ-,  join  T'O,  Ot  and  RS. 

Since  the  figure  is  symmetrical  with  respect  to  MN,  we  have  RT' 
=  RT-,  and  since  RT '=  OS,  it  follows  that  RT'  =  OS;  therefore, 


BOOK     I.  51 

RT' OS  is  a  parallelogram  (108),  and  RS  is  equal  and  parallel 
to  OT'. 

Again,  since  the  figure  is  symmetrical  with  respect  to  PQ,  we  have 
St  =  ST  =  OR;  therefore,  SROt  is  a  parallelogram,  and  RS  is 
equal  and  parallel  to  Ot.  Hence,  T,  0  and  t,  are  in  the  same 
straight  line,  since  there  can  be  but  one  parallel  to  RS  drawn 
through  the  same  point  0. 

Now  we  have  OT'  =  RS  and  Ot  =  RS,  and  consequently  OT'  = 
Ot\  therefore,  any  straight  line  T'Ot,  drawn  through  0,  is  bisected 
at  0 ;  that  is,  0  is  the  centre  of  symmetry  of  the  figure. 


BOOK   II. 

THE  CIRCLE. 

1.  DEFINITIONS.  A  circle  is  a  portion  of  a  plane  bounded  by  a 
curve,  all  the  points  of  which  are  equally  distant  from  a  point  within 
it  called  the  centre. 

The  curve  which  bounds  the  circle  is  called 
its  circumference-. 

Any  straight  line  drawn  from  the  centre 
to  the  circumference  is  called  a  radius. 

Any  straight  line  drawn  through  the  centre 
and  terminated  each  way  by  the  circumfer 
ence  is  called  a  diameter. 

In  the  figure,  0  is  the  centre,  and  the  curve  ABCEA  is  the  cir 
cumference  of  the  circle ;  the  circle  is  the  space  included  within  the 
circumference;  OA,  OB,  OC,  are  radii;  A OC is  a  diameter. 

By  the  definition  of  a  circle,  all  its  radii  are  equal ;  also  all  its 
diameters  are  equal,  each  being  double  the  radius.  r' 

If  one  extremity,  0,  of  a  line  OA  is  fixed,  while  the  line  revolves 
in  a  plane,  the  other  extremity,  A,  will  describe  a  circumference, 
whose  radii  are  all  equal  to  OA. 

2.  Definitions.  An  arc  of  a  circle  is  any  portion  of  its  circumfer 
ence  ;  as  DEF. 

A  chord  is  any  straight  line  joining  two  points  of  the  circum 
ference;  as  DF.  The  arc  DEF  is  said  to  be  subtended  by  its 
chord  DF. 

Every  chord  subtends  two  arcs,  which  together  make  up  the  whole 
circumference.  Thus  DF  subtends  both  the  arc  DEF  and  the  arc 

DCBAF.    When  an  arc  and  its  chord  are  spoken  of,  the  arc  less  than 
52 


BOOK     II.  53 

a  semi-circumference,  as  DEF,  is  always  understood,  unless  otherwise 
stated. 

A  segment  is  a  portion  of  the  circle  included  between  an  arc  and 
its  chord ;  thus,  by  the  segment  DEF  is  meant  the  space  included 
between  the  arc  DF  and  its  chord. 

A  sector  is  the  space  included  between  an  arc  and  the  two  radii 
drawn  to  its  extremities ;  as  A  OB. 

3.  From  the  definition  of  a  circle  it  follows  that  every  point 
within  the  circle  is  at  a  distance  from  the  centre  which  is  less  than 
the  radius ;  and  every  point  without  the  circle  is  at  a  distance  from 
the  centre  which  is  greater  than  the  radius.     Hence  (I.  40),  the 
locus  of  all  the  points  in  a  plane  which  are  at  a  given  distance  from  a 
given  point  is  the  circumference  of  a  circle  described  with  the  given  point 
as  a  centre  and  with  the  given  distance  as  a  radius. 

4.  It  is  also  a  consequence  of  the  definition  of  a  circle,  that  two 
circles  are  equal  when  the  radius  of  one  is  equal  to  the  radius  -of  the 
other,  or  when  (as  we  usually  say)  they  have  the  same  radius.     For 
if  one  circle  be  superposed  upon  the  other  so  that  their  centres  coin 
cide,  their  circumferences  will  coincide,  since  all  the  points  of  both 
are  at  the  same  distance  from  the  centre. 

If  when  superposed  the  second  circle  is  made  to  turn  upon  its 
centre  as  upon  a  pivot,  it  must  continue  to  coincide  with  the  first. 

5.  Postulate.  A  circumference  may  be  described  with  any  point  as 
a  centre  and  any  distance  as  a  radius. 


AKCS    AND   CHOEDS. 

PKOPOSITION  I.— THEOKEM. 

6.  A  straight  line  cannot  intersect  a  circumference  in  more  than  tivo 
points. 

For,  if  it  could  intersect  it  in  three  points,  the  three  radii  drawn 
to  these  three  points  would  be  three  equal  straight  lines  drawn  from 
the  same  point  to  the  same  straight  line,  which  is  impossible  (I.  36). 


54 


GEOMETRY. 


PROPOSITION  II.— THEOREM. 

7.  Every  diameter  bisects  the  circle  and  its  circumference. 
Let  AMBN  be  a  circle  whose  centre  is  0 ; 

then,  any  diameter  AOB  bisects  the  circle  and 
its  circumference. 

For,  if  the  figure  ANB  be  turned  about  AB 
as  an  axis  and  superposed  upon  the  figure 
AMB,  the  curve  ANB  will  coincide  with  the 
curve  AMB,  since  all  the  points  of  both  are 
equally  distant  from  the  centre.  The  two 

figures  then  coincide  throughout,  and  are  therefore  equal  in  all 
respects.  Therefore,^4J5  divides  both  the  circle  and  its  circumference 
into  equal  parts. 

8.  Definitions.  A  segment  equal  to  one  half  the  circle,  as  the  seg 
ment  AMB,  is  called  a  semi-circle.     An  arc  equal  to  half  a  circum 
ference,  as  the  arc  AMB,  is  called  a  semi-circumference. 


PROPOSITION  III.— THEOREM. 

9.  A  diameter  is  greater  than  any  other  chord. 

Let  A  C  be  any  chord  which  is  not  a  diame 
ter,  and  A  OB  a  diameter  drawn  through  A : 
then^!J5>  AC. 

For,  join  OC.  Then,  AO  -f-  00  >  AC 
(I.  66) ;  that  is,  since  all  the  radii  are  equal, 
AO+  OB>AC,orAB>Aa 


PROPOSITION  IV— THEOREM. 

10.  In  equal  circles,  or  in  the  same  circle,  equal  angles  at  the  centre 
intercept  equal  arcs  on  the  circumference,  and  conversely. 

Let  0,  0',  be  the  centre  of  equal 
circles,  and  AOB,  A'O'B',  equal  angles 
at  these  centres ;  then,  the  intercepted 
arcs,  AB,  A'B',  are  equal.  For,  one  of 
the  angles,  together  with  its  arc,  may  be 
superposed  upon  the  other;  and  when 


BOOK     II. 


55 


the  equal  angles  coincide,  their  intercepted  arcs  will  evidently  coin 
cide  also. 

Conversely,  if  the  arcs  AB,  A'B'  are  equal,  the  angles  A  OB, 
A'  O'B'  are  equal.  For,  when  one  of  the  arcs  is  superposed  upon  its 
equal,  the  corresponding  angles  at  the  centre  will  evidently  coincide. 

If  the  angles  are  in  the  same  circle,  the  demonstration  is  similar. 

11.  Definition.  A  fourth  part  of  a  circum 
ference  is  called  a  quadrant.  It  is  evident  from 
the  preceding  theorem  that  a  right  angle  at  the 
centre  intercepts  a  quadrant  on  the  circum 
ference. 

Thus,  two  perpendicular  diameters,  AOC, 
BOD,  divide  the  circumference  into  four  quad 
rants,  AB,  BC,  CD,  DA. 


PKOPOSIT10N  V.— THEOEEM. 

12.  In  equal  circles,  or  in  the  same  circle,  equal  arcs  are  subtended 
by  equal  chords,  and  conversely. 

Let  0,   O',be  the  centres  of  equal  circles,  and  AB,  A'B',  equal 
arcs;  then,  the  chords  AB,  A'B',  are 
equal. 

For,  drawing  the  radii  to  the  extremi 
ties  of  the  arcs,  the  angles   0  and   0'       V         °        I     \        of 
are  equal   (10),  and   consequently  the 
triangles    A  OB,    A  O'B',    are    equal 
(I.  76).     Therefore,  AB  =  AB'. 

Conversely,  if  the  chords  AB,  A'B',  are  equal,  the  triangles  AOB, 
A  O'B'  are  equal  (I.  80),  and  the  angles  0,  0'  are  equal.  There 
fore  (10),  arc  AB  =  arc  A'B'. 

If  the  arcs  are  in  the  same  circle,  the  demonstration  is  similar. 


PROPOSITION  VI.— THEOEEM. 

13.  In  equal  circles,  or  in  the  same  circle,  the  greater  arc  is  subtended 
by  the  greater  chord,  and  conversely ;  the  arcs  being  both  less  than  a 
semi-circumference: 


56 


GEOMETRY. 


Let  the  arc  A  C  be  greater  than  the 
arc  AB;  then,  the  chord  AC  is  greater 
than  the  chord  AB. 

For,  draw  the  radii   OA,  OB,   OC. 
In  the  triangles  AOC,  A  OB,  the  angle 
AOC  is  obviously  greater  than  the  angle  A  OB;  therefore,  (I.  84), 
chord  A  C  >  chord  AB. 

Conversely,  if  chord  AC  >  chord  AB,  then,  arc  AC  >  arc  AB. 
For,  in  the  triangles  AOC,  A  OB,  the  side  AC  >  the  side  AB; 
therefore  (I.  85),  angle  AOC  >  angle  A  OB;  and  consequently, 
arc  A  C  >  arc  AB. 

14.  Scholium.  If  the  arcs  are  greater  than  a  semi-circurnference, 
the  contrary  is  true ;  that  is,  the  arc  AMB,  which  is  greater  than  the 
arc  AM  C,  is  subtended  by  the  less  chord ;  and  conversely. 


PROPOSITION  VII.— THEOREM. 

15.  The  diameter  perpendicular  to  a  chord  bisects  the  chord  and  the 
arcs  subtended  by  it. 

Let  the  diameter  DOD'  be  perpendicular  to 
the  chord  AB  at  C;  then,  1st,  it  bisects  the 
chord.  For,  the  radii  OA,  OB  being  equal 
oblique  lines  from  the  point  0  to  the  line  AB, 
cut  off  equal  distances  from  the  foot  of  the  per 
pendicular  (I.  36);  therefore,  AC=BC. 

2d.  The  subtended  arcs  ADB,  AD'B,  are 
bisected  at  D  and  Dr,  respectively.  For,  every  point  in  the  per 
pendicular  DOD'  drawn  at  the  middle  of  AB  being  equally  distant 
from  its  extremities  A  and  B  (I.  38),  the  chords  AD  and  BD  are 
equal;  therefore,  (12),  the  arcs  AD  and  BD  are  equal.  For  the 
same  reason,  the  arcs  AD'  and  BD'  are  equal. 

16.  Corollary  I.  The  perpendicular  erected  upon  the  middle  of  a 
chord    passes   through    the  centre  of  the  circle,  and  through   the 
middle  of  the  arc  subtended  by  the  chord. 

Also,  the  straight  line  drawn  through  any  two  of  the  three  points 
0,  C,  D,  passes  through  the  third  and  is  perpendicular  to  the 
chord  AB. 


BOOK      II.  57 


17.  Corollary  II.  The  middle  points  of  any 
number  of  parallel  chords  all  lie  in  the  same 
diameter  perpendicular  to  the  chords. 

In  other  words,  the  locus  of  the  middle  points 
of  a  system  of  parallel  chords  is  the  diameter 
perpendicular  to  these  chords. 


PROPOSITION  VIII.— THEOREM. 

18.  In  the  same  circle,  or  in  equal  circles,  equal  chords  are  equally 
distant  from  the  centre ;  and  of  two  unequal  chords,  the  less  is  at  the 
greater  distance  from  the  centre. 

1st.  Let  AB,  CD,  be  equal  chords ;  OE, 
OF,  the  perpendiculars  which  measure  their 
distances  from  the  centre  0;  then,  OE  = 
OF. 

For,  since  the  perpendiculars  bisect  the 
chords  (15),  AE=CF;  hence  (I.  83),  the 
right  triangles  AOE  and  COF  are  equal, 
and  OE  =  OF. 

2d.  Let  CG,  AB,  be  unequal  chords;  OE,  OH,  their  distances 
from  the  centre  ;  and  let  CG  be  less  than  AB ;  then,  OH  >  OE. 

For,  since  chord  AB  >  chord  CG,  we  have  arc  AB  >  arc  CG ; 
so  that  if  from  C  we  draw  the  chord  CD  •=  AB,  its  subtended  arc 
CD,  being  equal  to  the  arc  AB,  will  be  greater  than  the  arc  CG. 
Therefore  the  perpendicular  OH  will  intersect  the  chord  CD  in  some 
point  J.  Drawing  the  perpendicular  OF  to  CD,  we  have,  by  the 
first  part  of  the  demonstration,  OF  =  OE.  But  OH  >  01,  and 
01 >  OF  (1.  28);  still  more,  then,  is  OH>  OF,  or  OH>  OE. 

If  the  chords  be  taken  in  two  equal  circles,  the  demonstration  is 
the  same. 

19.  Corollary  I.  The  converse  of  the  proposition  is  also  evidently 
true,  namely :   in  the  same  circle,  or  in  equal  circles,  chords  equally 
distant  from  the  centre  are  equal;  and  of  two  chords  unequally  distant 
from  the  centre,  that  is  the  greater  whose  distance  from  the  centre  is 
the  less. 


f>8 


GEOMETRY. 


20.  Corollary  II.  The  least  chord  that  can  be 
drawn  in  a  circle  through  a  given  point  P  is  the 
chord,  AB,  perpendicular  to  the  line  OP  joining 
the  given  point  and  the  centre.  For,  if  CD  is 
any  other  chord  drawn  through  P,  the  perpen 
dicular  OQ  to  this  chord  is  less  than  OP;  there 
fore,  by  the  preceding  corollary,  CD  is  greater 
than  AB. 


PKOPOSITION  IX.— THEOREM. 

21.  Through  any  three  points,  not  in  the  same  straight  line,  a  circum 
ference  can  be  made  to  pass,  and  but  one. 

Let  A,  B,  C,  be  any  three  points  not  in  the 
same  straight  line. 

1st.  A  circumference  can  be  made  to  pass 
through  these  points.  For,  since  they  are 
not  in  the  same  straight  line,  the  lines  AB, 
BC,  AC,  joining  them  two  and  two,  form  a 
triangle,  and  the  three  perpendiculars  DE, 

FG,  HK,  erected  at  the  middle  points  of  the  sides,  meet  in  a  point 
0  which  is  equally  distant  from  the  three  points  A,  B,  C,  (I.  131). 
Therefore  a  circumference  described  from  0  as  a  centre  and  a  radius 
equal  to  any  one  of  the  three  equal  distances  OA,  OB,  OC,  will  pass 
through  the  three  given  points. 

2d.  Only  one  circumference  can  be  made  to  pass  through  these 
points.  For  the  centre  of  any  circumference  passing  through  the 
three  points  must  be  at  once  in  two  perpendiculars,  as  DE,  FG,  and 
therefore  at  their  intersection ;  but  two  straight  lines  intersect  in 
only  one  point,  and  hence  0  is  the  centre  of  the  only  circumference 
that  can  pass  through  the  three  points. 

22.  Corollary.  Two  circumferences  can  intersect  in  but  two  points; 
for,  they  could  not  have  a  third  point  in  common  without  having  the 
same  centre  and  becoming  in  fact  but  one  circumference. 


BOOK     II. 


59 


TANGENTS  AND  SECANTS. 

23.  Definitions.  A  tangent  is  an  indefinite  straight  line  which  has 
but  one  point  in  common  with  the  cir 
cumference;    as   ACB.     The   common 

point,  O,  is  called  the  point  of  contact, 
or  the  point  of  tangency.  The  circum 
ference  is  also  said  to  be  tangent  to  the 
line  AB  at  the  point  C. 

A  secant  is  a  straight  line  which 
meets  the  circumference  in  two  points ; 
as  EF. 

24.  Definition.  A  rectilinear  figure  is  said  to 
be  circumscribed  about  a  circle  when  all  its  sides 
are  tangents  to  the  circumference. 

In  the  same  case,  the  circle  is  said  to  be  in 
scribed  in  the  figure. 


PKOPOSITION  X.— THEOEEM. 

25.  A  straight  line  oblique  to  a  radius  at  its  extremity  cuts  the  cir 
cumference. 

Let  AB  be  oblique  to  the  radius  0  C  at  its 
extremity  0;  then,  AB  cuts  the  circumfer 
ence  at  C,  and  also  in  a  second  point  D. 

For,  let  OE  be  the  perpendicular  from  0 
upon  AB-,  then  OE  <  00,  and  the  point  E 
is  within  the  circumference.  Therefore  AB 
cuts  the  circumference  in  0,  and  must  evi 
dently  cut  it  in  a  second  point  D. 


60 


GEOMETRY. 


PROPOSITION  XL— THEOREM. 

26.  A  straight  line  perpendicular  to  a  radius  at  ite  extremity  is  a 
tangent  to  the  circle. 

Let  AS  be  perpendicular  to  the  radius  OC 
at  its  extremity  C;  then,  AB  is  a  tangent  to 
the  circle  at  the  point  C. 

For,  from  the  centre  0  draw  the  oblique 
line  OD  to  any  point  of  AB  except  C.  Then, 
OD  >  OC,  and  D  is  a  point  without  the  cir 
cumference.  Therefore  AB  having  all  its 
points  except  C  without  the  circumference,  has  but  the  point  C  in 
common  with  it,  and  is  a  tangent  at  that  point  (23). 

27.  Corollary.  Conversely,  a  tangent  AB  at  any  point  C  is  perpen 
dicular  to  the  radius  OC  drawn  to  that  point.     For,  if  it  were  not 
perpendicular  to  the  radius  it  would  cut  the  circumference  (25),  and 
would  not  be  a  tangent. 

28.  Scholium.  If  a  secant  EF,  passing  through  a  point  C  of  the 
circumference,  be  supposed    to   revolve 

upon  this  point,  as  upon  a  pivot,  its 
second  point  of  intersection,  D,  will 
move  along  the  circumference  and  ap 
proach  nearer  and  nearer  to  C.  When 
the  second  point  comes  into  coincidence 
with  C,  the  revolving  line  ceases  to  be 
strictly  a  secant,  and  becomes  the  tan 
gent  AB;  but,  continuing  the  revolution, 

the  revolving  line  again  becomes  a  secant,  as  E'F',  and  the  second 
point  of  intersection  reappears  on  the  other  side  of  (7,  as  at  D'. 

If,  then,  our  revolving  line  be  required  to  be  a  secant  in  the  strict 
sense  imposed  by  our  definition,  that  is  a  line  meeting  the  circum 
ference  in  two  points,  this  condition  can  be  satisfied  only  by  keeping 
the  second  point  of  intersection,  D,  distinct  from  the  first  point,  (7, 
however  near  these  points  may  be  brought  to  each  other ;  and,  there 
fore,  under  this  condition,  the  tangent  is  often  called  the  limit  of  the 
secants  drawn  through  the  point  of  contact ;  that  is  to  say,  a  limit 
toward  which  the  secant  continually  approaches,  as  the  second  point 


BOOK    II.  61 

of  intersection  (on  either  side  of  the  first)  continually  approaches 
the  first,  but  a  limit  which  is  never  reached  by  the  secant  as  such. 

On  the  other  hand,  as  the  tangent  is  but  one  of  the  positions  of 
our  revolving  line,  it  has  properties  in  common  with  the  secant;  and 
in  order  to  exhibit  such  common  properties  in  the  most  striking 
manner,  it  is  often  expedient  to  regard  the  tangent  as  a  secant  whose 
two  points  of  intersection  are  coincident.  But  it  is  to  be  observed  that 
we  then  no  longer  consider  the  secant  as  a  cutting  line,  but  simply  as 
a  line  drawn  through  two  points  of  the  curve ;  and  we  include  the 
tangent  as  that  special  case  of  such  a  line  in  which  the  two  points 
are  coincident.  In  this,  we  generalize  in  the  same  way  as  in  algebra, 
when  we  say  that  the  expression  x  —  a  —  b  signifies  that  x  is  the 
difference  of  a  and  b,  even  when  a  =  b,  and  there  is  really  no  differ 
ence  between  a  and  b. 


PKOPOSITION  XII.— THEOREM. 

29.   Two  parallels  intercept  equal  arcs  on  a  circumference. 

We  may  have  three  cases : 

1st.  When  the  parallels  AB,  CD,  are  both       — ^ 

secants  ;  then,  the  intercepted  arcs  AC  and  BD 
are  equal.  For,  let  OM  be  the  radius  drawn 
perpendicular  to  the  parallels.  By  Prop.  VII. 
the  point  M  is  at  once  the  middle  of  the  arc 


AMB  and  of  the  arc  CMD,  and  hence  we  have       G  N  H 

AM  =  BM  and  CM  =  DM, 
whence,  by  subtraction, 

AM  —CM  =BM— 

that  is, 

=  BD. 


2d.  When  one  of  the  parallels  is  a  secant,  as  AB,  and  the  other  is 
a  tangent,  as  EF  at  M,  then,  the  intercepted  arcs  AM  and  BM  are 
equal.  For,  the  radius  OM  drawn  to  the  point  of  contact  is  per 
pendicular  to  the  tangent  (27),  and  consequently  perpendicular  also 
to  its  parallel  AB\  therefore,  by  Prop.  VII.,  AM  =  BM. 

3d.  When  both  the  parallels  are  tangents,  as  EF  at  M,  and  GH 


62  GEOMETRY. 

at  N;  then,  the  intercepted  arcs  MAN  and  MBN  are  equal.     For, 

drawing  any  secant  AB  parallel  to  the  tangents, 

we  have  by  the  second  case,  ^ — — =   M 


AM  =  BM  and  AN  =  BN, 
whence,  by  addition, 

AM+AN  =  BM+BN, 


that  is, 


MAN  =  MBN; 


JV  H 


and  each  of  the  intercepted  arcs  in  this  case  is  a  serai-circumference. 

30.  Scholium  1.  The  straight  line  joining  the  points  of  contact  of 
two  parallel  tangents  is  a  diameter. 

31.  Scholium  2.  According  to  the  principle  of  (28),  the  tangent 
being  regarded  as  a  secant  whose  two  points  of  intersection  are  coin 
cident,  the  demonstration  of  the  first  case  in  the  preceding  theorem 
embraces  that  of  the  other  two  cases. 


RELATIVE  POSITION  OF  TWO  CIRCLES. 

32.  Definition.   Two  circles  are  eoncentric,  when   they  have  the 
same  centre. 

33.  Definition.  Two  circumferences  are  tangent  to  each  other,  or 
touch  each  other,  when  they  have  but  one  point  in  common.     The 
common  point  is  called  the  point  of  contact,  or  the  point  of  tangency. 

Two  kinds  of  contact  are  distinguished :  external  contact,  when 
each  circle  is  outside  the  other ;  internal  contact,  when  one  circle  is 
within  the  other. 


PROPOSITION  XIII.— THEOREM. 

34.    When  two  circumferences  intersect,  the  straight  line  joining  their 
centres  bisects  their  common  chord  at  right  angles. 

Let  0  and  0'  be  the  centres  of  two 
circumferences  which  intersect  in  the 
points  A,  B;  then,  the  straight  line  00' 
bisects  their  common  chord  AB  at  right 
angles. 

For,  the  perpendicular  to  AB  erected 


BOOK     II. 


63 


at  its  middle  point  C,  passes  through  both  centres  (16) ;  and  there 
can  be  but  one  straight  line  drawn  between  the  two  points  0  and  0'. 
35.  Corollary.  When  two  circumferences  are  tangent  to  each  other, 
their  point  of  contact  is  in  the  straight  line  joining  their  centres.  It 
has  just  been  proved  that  when  two  circumferences  intersect,  the  two 
points  of  intersection  lie  at  equal  distances  from  the  line  joining  the 
centres  and  on  opposite  sides  of  this  line.  Now  let  the  circles  be 
supposed  to  be  moved  so  as  to  cause  the  points  of  intersection  to 
approach  each  other;  these  points  will 
ultimately  come  together  on  the  line 
joining  the  centres,  and  be  blended  in  a 
single  point  C,  common  to  the  two  cir 
cumferences,  which  will  then  be  their 
point  of  contact.  The  perpendicular  to 
00'  erected  at  C  will  then  be  a  common 

tangent  to  the  two  circumferences  and  take  the  place  of  the  common 
chord. 


PBOPOSITION  XIV— THEOEEM. 

36.    When  two  circumferences  are  wholly  exterior  to  each  other,  the 
distance  of  their  centres  is  greater  than  the  sum  of  their  radii. 

Let  0,  0'  be  the  centres.     Their  dis 
tance  00'  is  greater  than  the  sum  of 
the  radii  OA,  O'B,  by  the  portion  AB      \         0- 
interposed  between  the  circles. 


PEOPOSITION  XV— THEOEEM. 

37.    When  two  circumferences  are  tangent  to  each  other  externally,  the 
distance  of  their  centres  is  equal  to  the  sum  of  their  radii. 

Let  0,  0',  be  the  centres,  and  C  the  point 
of  contact.     The  point  C  being  in  the  line 
joining  the  centres  (35),  we  have  00'  —     |         o- 
OC  +  O'C. 


0' 


64 


GEOMETRY. 


PROPOSITION  XVI.— THEOREM. 

38.  When  two  circumferences  intersect,  the  distance  of  their  centres 
is  less  than  the  sum  of  their  radii  and  greater  than  the  difference  of  their 
radii. 

Let  0  and  0'  be  their  centres,  and  A 
one  of  their  points  of  intersection.  The 
point  A  is  not  in  the  line  joining  the 
centres  (34)  ;  and  consequently  there  is 
formed  the  triangle  A 00',  in  which  we 
have  00'  <  OA  +  O'A,  and  also 
00'  >  OA  —  O'A  (I.  67). 


PROPOSITION  XVII— THEOREM. 

39.    When  two  circumferences  are  tangent  to  each  other  internally, 
the  distance  of  their  centres  is  equal  to  the  difference  of  their  radii. 


Let  0,  0',  be  the  centres,  and  (7  the  point  of 
contact.  The  point  C  being  in  the  line  joining 
the  centres  (35),  we  have  00'  =00—  OfC. 


PROPOSITION  XVIII.— THEOREM. 

40.    When  one  circumference  is  wholly  within  another,  the  distance 
of  their  centres  is  less  than  the  difference  of  their  radii. 


Let  0,  0',  be  the  centres.  We  have  the  dif 
ference  of  the  radii  OA  —  O'B  =  00'  -f  AR 
Hence  00'  is  less  than  the  difference  of  the 
radii  by  the  distance  AB. 


41.  Corollary.  The  converse  of  each  of  the  preceding  five  propo 
sitions  is  also  true :  namely — 

1st.  When  the  distance  of  the  centres  is  greater  than  the  sum  of 
the  radii,  the  circumferences  are  wholly  exterior  to  each  other. 


B  O  O  K     1 1 .  65 

2d.  When  the  distance  of  the  centres  is  equal  to  the  sum  of  the 
radii,  the  circumferences  touch  each  other  externally. 

3d.  When  the  distance  of  the  centres  is  less  than  the  sum  of  the 
radii,  but  greater  than  their  difference,  the  circumferences  intersect. 

4th.  When  the  distance  of  the  centres  is  equal  to  the  difference 
of  the  radii,  the  circumferences  touch  each  other  internally. 

5th.  When  the  distance  of  the  centres  is  less  than  the  difference 
of  the  radii,  one  circumference  is  wholly  within  the  other. 

MEASUKE  OF  ANGLES. 

As  the  measurement  of  magnitude  is  one  of  the  principal  objects 
of  geometry,  it  will  be  proper  to  premise  here  some  principles  in 
regard  to  the  measurement  of  quantity  in  general. 

42.  Definition.  To  measure  a  quantity  of  any  kind  is  to  find  how 
many  times- it  contains  another  quantity  of  the  same  kind  called  the 
unit. 

Thus,  to  measure  a  line  is  to  find  the  number  expressing  how  many 
times  it  contains  another  line  called  the  unit  of  length,  or  the  linear 
unit. 

The  number  which  expresses  how  many  times  a  quantity  contains 
the  unit  is  called  the  numerical  measure  of  that  quantity. 

43.  Definition.  The  ratio  of  two  quantities  is  the  quotient  arising 

^ 
from  dividing  one  by  the  other ;  thus,  the  ratio  of  A  to  B  is  — 

B 

To  find  the  ratio  of  one  quantity  to  another  is,  then,  to  find  how 
many  times  the  first  contains  the  second ;  therefore,  it  is  the  same 
thing  as  to  measure  the  first  by  the  second  taken  as  the  unit  (42). 
It  is  implied  in  the  definition  of  ratio,  that  the  quantities  compared 
are  of  the  same  kind. 

Hence,  also,  instead  of  the  definition  (42),  we  may  say  that  to 
measure  a  quantity  is  to  find  its  ratio  to  the  unit. 

The   ratio  of  two  quantities  is  the  same  as  the  ratio  of  their 
numerical  measures.     Thus,  if  P  denotes  the  unit,  and  if  P  is  con 
tained  m  times  in  A  and  n  times  in  B,  then, 
•. 

A mP m 

B    •    nP  • ,"  n 

44.  Definition.  Two  quantities  are  commensurable  when  there  is 


66  GEOMETRY. 

some  third  quantity  of  the  same  kind  which  is  contained  a  whole 
number  of  times  in  each.  This  third  quantity  is  called  the  common 
measure  of  the  proposed  quantities. 

Thus,  the  two  lines,  A  and  B,  are  commensurable,  if  there  is  some 
line,  C,  which  is  contained  a  whole  num 
ber   of  times   in   each,  as,  for   example, 
7  times  in  A,  and  4  times  in  B.  s> — i — « — • — « 

The  ratio  of  two  commensurable  quan-         Cl , 

titles  can,  therefore,  be  exactly  expressed 

by  a  number  whole  or  fractional  (as  in   the  preceding  example 

7\ 
by  -),  and  is  called  a  commensurable  ratio. 

v 

45.  Definition.   Two   quantities   are   incommensurable   when   they 
have  no  common  measure.     The  ratio  of  two  such  quantities  is  called 
an  incommensurable  ratio. 

If  A  and  B  are  two  incommensurable  quantities,  their  ratio  is  still 

expressed  by  — 

46.  Problem.   To  find  the  greatest  common  measure  of  two  quantities. 
The  well-known  arithmetical  process  may  be  extended  to  quantities 
of  all  kinds.     Thus,  suppose  AB  and  CD  are  two  straight  lines 
whose  common  measure  is  required.     Their  greatest  common  meas 
ure  cannot  be  greater  than  the  less  line 

CD.     Therefore,  let  CD  be  applied  to  AB      Al         ' ' I" B 
as  many  times  as  possible,  suppose  3  times,        Cr~r~pD 
with  a  remainder  EB  less  than  CD.     Any 

common  measure  of  AB  and  CD  must  also  be  a  common  measure 
of  CD  and  EB ;  for  it  will  be  contained  a  whole  number  of  times  in 
CD,  and  in  AE,  which  is  a  multiple  of  CD,  and  therefore  to  measure 
AB  it  must  also  measure  the  part  EB.  Hence,  the  greatest  common 
measure  of  AB  and  CD  must  also  be  the  greatest  common  measure 
of  CD  and  EB.  This  greatest  common  measure  of  CD  and  EB 
cannot  be  greater  than  the  less  line  EB;  therefore,  let  EB  be  applied 
as  many  times  as  possible  to  CD,  suppose  twice,  with  a  remainder 
FD.  Then,  by  the  same  reasoning,  the  greatest  common  measure 
of  CD  and  EB,  and  consequently  also  that  of  AB  and  CD,  is  the 
greatest  common  measure  of  EB  and  FD.  Therefore,  let  FD  be 
applied  to  EB  as  many  times  as  possible :  suppose  it  is  contained 


B  O  O  K     I  1 .  67 

exactly  twice  in  EB  without  remainder ;  the  process  is  then  com 
pleted,  and  we  have  found  FD  as  the  required  greatest  common 
measure. 

The  measure  of  each  line,  referred  to  FD  as  the  unit,  will  then  be 
as  follows :  we  have 

EB  =  2FD, 

CD  =  2EB  +  FD  =  ±FD  +  FD  =  5FD, 
AB  =  BCD  +  EB  =  15FD  +  2FD  =  17FD. 
The  proposed  lines  are  therefore  numerically  expressed,  in  terms  of 

the  unit  FD,  by  the  numbers  17  and  5 ;  and  their  ratio  is  — 

5 

47.  When  the  preceding  process  is  applied  to  two  quantities  and 
no  remainder  can  be  found  which  is  exactly  contained  in  a  pre 
ceding  remainder,  however  far  the  process  be  continued,  the  two 
quantities  have  no  common  measure;  that  is,  they  are  incommen 
surable,  and  their  ratio  cannot  be  exactly  expressed  by  any  number 
whole  or  fractional. 

48.  But  although   an   incommensurable  ratio  cannot  be  exactly 
expressed  by  a  number,  it  may  be  approximately  expressed  by  a 
number  within  any  assigned  measure  of  precision. 

Suppose  —  denotes  the  incommensurable  ratio  of  two  quantities 
B 

A  and  B ;  and  let  it  be  proposed  to  obtain  an  approximate  numeri 
cal  expression  of  this  ratio  that  shall  be  correct  within  an  assigned 

measure  of  precision,  say Let  B  be  divided  into  100  equal 

parts,  and  suppose  A  is  found  to  contain  314  of  these  parts  with  a 
remainder  less  than  one  of  the  parts ;  then,  evidently,  we  have 

A       314     ...      1 

—  = within > 

B       100  100 

that  is, is  an  approximate  value  of  the  ratio  —  within  the  as 
signed  measure  of  precision. 

A 

To  generalize  this,  —  denoting  as  before  the  incommensurable 
B 

ratio  of  the  two  quantities  A  and  B,  let  B  be  divided  into  n  equal 


68  GEOMETRY. 

parts,  and  let  A  contain  m  of  these  parts  with  a  remainder  less  than 
one  of  the  parts  ;  then  we  have 

A       m     ...     1 
—  =  -  within  -  ; 
B        n  n 

and,  since  n  may  be  taken  as  great  as  we  please,  -  may  be  made  less 

ft 

than  any  assigned  measure  of  precision,  and  —  will  be  the  approxi- 

n 

j^ 
mate  value  of  the  ratio  —  within  that  assigned  measure. 

B 

49.  Theorem.  Two  incommensurable  ratios  are  equal,  if  their  approxi 
mate  numerical  values  are  always  equal,  when  both  are  expressed  within 
the  same  measure  of  precision  however  small. 

A          Af 

Let  —  and  —  f  be  two  incommensurable  ratios  whose  approximate 
IB          B 

numerical  values  are  always  the  same  when  the  same  measure  of 
precision  is  employed  in  expressing  both  ;  then,  we  say  that 


For,  let  -  be  any  assumed  measure  of  precision,  and  in  accordance 

with  the  hypothesis  of  the  theorem,  suppose  that  for  any  value  of 

1  A    A' 

->the  ratios  —  »  —  have  the  same  approximate  numerical  expres- 

n  B    B' 

siou,  say  —  >  each   ratio    exceeding  —  by  a  quantity  less  than  -; 
n  n  n 

then,  these  ratios  cannot  differ  from  each  other  by  so  much  as  -• 

But  the  measure  -  may  be  assumed  as  small  as  we  please,  that  is  less 

A         A' 
than  any  assignable  quantity  however  small  ;  hence  —  and  —  cannot 

differ  by  any  assignable  quantity  however  small,  and  therefore  they 
must  be  equal. 

The  student  should  study  this  demonstration  in  connection  with 
that  of  Proposition  XIX.,  which  follows. 


BOOK     II. 


50.  Definition.  A  proportion  is  an  equality  of  ratios.     Thus,  if  the 

ratio  —  is  equal  to  the  ratio  —>  the  equality 
B  B 


B~  B' 

is  a  proportion.     It  may  be  read  :  "  Ratio  of  A  to  B  equals  ratio  of 
A'  to  B',"  or  "A  is  to  B  as  A'  is  to  B'." 
A  proportion  is  often  written  as  follows  : 

A  :  B  =  A  :  B' 

where  the  notation  A  :  B  is  equivalent  to  A  -j-  B.  When  thus 
written,  A  and  B'  are  called  the  extremes,  B  and  A'  the  means,  and 
B'  is  called  a  fourth  proportional  to  A,  B  and  A'  ;  the  first  terms 
A  and  JL',  of  the  ratios  are  called  the  antecedents  —  the  second  terms, 
B  and  B',  the  consequents. 

When  the  means  are  equal,  as  in  the  proportion 

A  :  B  =  B  :  C, 

the  middle  term  B  is  called  a  mean  proportional  between  A  and  (7, 
and  C  is  called  a  third  proportional  to  A  and  B. 

PKOPOSITION  XIX.—  THEOEEM. 

51.  In  the  same  circle,  or  in  equal  circles,  two  angles  at  the  centre  are 
in  the  same  ratio  as  their  intercepted  arcs. 

Let  A  OB  and  AOC  be  two  angles  at  the  centre  of  the  same,  or  at 
the  centres  of  equal  circles;  AB  and  AC,  their  intercepted  arcs; 
then, 

AOB 


AOC~  AC 

1st.  Suppose  the  arcs  to  have 
a  common  measure  which  is  con 
tained,  for  example,  7  times  in 

the  arc  AB  and  4  times  in  the  arc  AC;  so  that  if  AB  is  divided 
into  7  parts,  each  equal  to  the  common  measure,  AC  will  contain  4 
of  these  parts.  Then  the  ratio  of  the  arcs  AB  and  A  C  is  7  :  4  ; 
that  is, 


70 


GEOMETRY. 


AB 

AC 


Drawing   radii    to    the   several      At 

points  of  division  of  the  arcs, 

the  partial  angles  at  the  centre 

subtended  by  the  equal  partial  arcs  will  be  equal  (10) ;   therefore 

the  angle  AOB  will  be  divided  into  7  equal  parts,  of  which  the 

angle  AOC  will  contain  4;  hence  the  ratio  of  the  angles  AOB  and 

A OC is  7:4;  that  is, 

AOB       7 
4 


Therefore,  we  have 


or, 


AOC 


AOB 


AB 
AC 


AOC 
AOB:AOC=AB:AC. 


2d.  If  the  arcs  are  incommensurable,  suppose  one  of  them,  as  AC, 
to  be  divided  into  any  number  n  of  equal  parts ;  then  AB  will  con 
tain  a  certain  number  m  of  these  parts,  plus  a  remainder  less  than 

one  of  these  parts.     The  numerical  expression  of  the  ratio will 

AC 

then  be  -,  correct  within  -  (48).     Drawing  radii  to  the  several 

n  n  ^     } 

points  of  division  of  the  arcs,  the  angle  A  OC  will  be  divided  into  n 
equal  parts,  and  the  angle  A  OB  will  contain  m  such  parts,  plus  a 
remainder  less  than  one  of  the  parts.  Therefore,  the  numerical 

expression  of  the  ratio will  also  be  —  >  correct  within  -. ;  that 


AOC 


n 


is,  the  ratio has  the  same  approximate  numerical  expression  as 

A  OC 

the  ratio  - — >  however  small  the  parts  into  which  A  C  is  divided  ; 
A  C> 

therefore  these  ratios  must  be  absolutely  equal  (49),  and  we  have  for 
incommensurable,  as  well  as  for  commensurable,  arcs, 

AOB      AB 


or, 


AOC      AC 
AOB:  AOC=AB:AC. 


BOOK    II.  71 


PKOPOSITION  XX.— THEOEEM. 

52.  The  numerical  measure  of  an  angle  at  the  centre  of  a  circle  is 
the  same  as  the  numerical  measure  of  its  intercepted  arc,  if  the  adopted 
unit  of  angle  is  the  angle  at  the  centre  which  intercepts  the  adopted  unit 
of  arc. 

Let  A  OB  be  an  angle  at  the  centre  0,  and 
AB  its  intercepted  arc.  Let  A  OC  be  the 
angle  which  is  adopted  as  the  unit  of  angle, 
and  let  its  intercepted  arc  AC  be  the  arc 
which  is  adopted  as  the  unit  of  arc.  By 
Proposition  XIX.  we  have 

AOB  _AB 
AOC~  AC 

But  the  first  of  these  ratios  is  the  measure  (42)  of  the  angle  A  OB 
referred  to  the  unit  AOC',  and  the  second  ratio  is  the  measure  of  the 
arc  AB  referred  to  the  unit  AC.  Therefore,  with  the  adopted  units, 
the  numerical  measure  of  the  angle  A  OB  is  the  same  as  that  of  the 
arc  AB. 

53.  Scholium  I.  This  theorem,  being  of  frequent  application,  is 
usually  more  briefly,  though  inaccurately,  expressed  by  saying  that 
an  angle  at  the  centre  is  measured  by  its  intercepted  arc.     In  this  con 
ventional  statement  of  the  theorem,  the  condition  that  the  adopted 
units  of  angle  and  arc  correspond  to  each  other  is  understood ;  and 
the  expression  "  is  measured  by"  is  used  for  "has  the  same  numerical 
measure  as." 

54.  Scholium  II.  The  right  angle  is,  by  its  nature,  the  most  simple 
unit  of  angle ;  nevertheless  custom  has  sanctioned  a  different  unit. 

The  unit  of  angle  generally  adopted  is  an  angle  equal  to  -^ih 
part  of  a  right  angle,  called  a  degree,  and  denoted  by  the  symbol  °. 
The  corresponding  unit  of  arc  is  -^th  part  of  a  quadrant  (11),  and 
is  also  called  a  degree. 

A  right  angle  and  a  quadrant  are  therefore  both  expressed  by  90°. 
Two  right  angles  and  a  semi-circumference  are  both  expressed  by 
180°.  Four  right  angles  and  a  whole  circumference  are  both  ex 
pressed  by  360°. 

The  degree  (either  of  angle  or  arc)  is  subdivided  into  minutes  and 


72 


GEOMETRY. 


seconds,  denoted  by  the  symbols  '  and  " :  a  minute  being  -g^th  part 
of  a  degree,  and  a  second  being  -g^th  part  of  a  minute.  Fractional 
parts  of  a  degree  less  than  one  second  are  expressed  by  decimal  parts 
of  a  second. 

An  angle,  or  an  arc,  of  any  magnitude  is,  then,  numerically  ex 
pressed  by  the  unit  degree  and  its  subdivisions.  Thus,  for  example, 
an  angle  equal  to  -fth  of  a  right  angle,  as  well  as  its  intercepted  arc, 
will  be  expressed  by  12°  51'  25".  714  .... 

55.  Definition.  When  the  sum  of  two  arcs  is  a  quadrant  (that  is, 
90°),  each  is  called  the  complement  of  the  other. 

When  the  sum  of  two  arcs  is  a  semi-circumference  (that  is,  180°), 
each  is  called  the  supplement  of  the  other.  See  (I.  18,  19). 

56.  Definitions.  An  inscribed  angle  is  one  whose  vertex  is  on  the 
circumference  and  whose  sides  are  chords ;  as  BA  C. 

In  general,  any  rectilinear  figure,  as  ABC,  is 
said  to  be  inscribed  in  a  circle,  when  its  angular 
points  are  on  the  circumference;  and  the  circle 
is  then  said  to  be  circumscribed  about  the  figure. 

An  angle  is  said  to  be  inscribed  in  a  segment 
when  its  vertex  is  in  the  arc  of  the  segment,  and 
its  sides  pass  through  the  extremities  of  the  sub 
tending  chord.     Thus,  the  angle  BA  C  is  inscribed  in  the  segment 
BAG. 


PROPOSITION  XXI.— THEOREM. 

57.  An  inscribed  angle  is  measured  by  one-half  its  intercepted  arc. 

There  may  be  three  cases : 

1st.  Let  one  of  the  sides  AB  of  the  inscribed 
angle  BAG  be  a  diameter;  then,  the  measure 
of  the  angle  BAG  is  one-half  the  arc  BC. 

For,  draw  the  radius  0(7.  Then,  AOC  being 
an  isosceles  triangle,  the  angles  OA C  and  OCA 
are  equal  (I.  86).  The  angle  BOG,  an  exterior 
angle  of  the  triangle  AOC,  is  equal  to  the  sum 
of  the  interior  angles  (Li (7 and  OCA  (I.  69),  and  therefore  double 


BOOK      II. 


73 


either  of  them.  But  the  angle  BOG,  at  the  centre,  is  measured  by 
the  arc  BC;  therefore,  the  angle  OAC  is  measured  by  one-half  the 
arc  BC. 

2d.  Let  the  centre  of  the  circle  fall  within  the  inscribed  angle 
BAG]  then,  the  measure  of  the  angle  BAG  is  one-half  of  the 
arc  B  C. 

For,  draw  the  diameter  AD.  The  measure  of 
the  angle  BAD  is,  by  the  first  case,  one-half  the 
arc  BD ;  and  the  measure  of  the  angle  CAD  is 
one-half  the  arc  CD',  therefore,  the  measure  of 
the  sum  of  the  angles  BAD  and  CAD  is  one-half 
the  sum  of  the  arcs  BD  and  CD;  that  is,  the 
measure  of  the  angle  BA  C  is  one-half  the  arc  BC. 

3d.  Let  the  centre  of  the  circle  fall  without  the  inscribed  angle 
BAG;  then,  the  measure  of  the  angle  BAG  is 
one-half  the  arc  BC. 

For,  draw  the  diameter  AD.  The  measure  of 
the  angle  BAD  is,  by  the  first  case,  one-half  the 
arc  BD ;  and  the  measure  of  the  angle  CAD  is 
one-half  the  arc  CD;  therefore,  the  measure  of 
the  difference  of  the  angles  BAD  and  CAD  is 
one-half  the  difference  of  the  arcs  BD  and  CD ; 
that  is,  the  measure  of  the  angle  BAG  is  one-half  the  arc  BC. 


58.  Corollary  I.  All  the  angles  BAG,  BDC, 
etc.,  inscribed  in  the  same  segment,  are  equal. 
For  each  is  measured  by  one-half  the  same 
arc  BMC. 


59.  Corollary  II.  Any  angle  BA  C}  inscribed  in 
a  semicircle  is  a  right  angle.  For  it  is  measured 
by  half  a  semi-circumference,  or  by  a  quad 
rant  (54). 

7 


74 


GEOMETRY. 


60.  Corollary  III.  Any  angle  B A  C,  inscribed 
in  a  segment  greater  than  a  semicircle,  is  acute ; 
for  it  is  measured  by  half  the  arc  BDC,  which 
is  less  than  a  semi-circumference. 

Any  angle  BDC,  inscribed  in  a  segment  less 
than  a  semicircle,  is  obtuse ;  for  it  is  measured 
by  half  the  arc  BA  C,  which  is  greater  than  a 
semi-circumference. 

61.  Corollary  IV.  The  opposite  angles  of  an  inscribed  quadrilateral 
ABDC,  are  supplements  of  each  other.     For  the  sum  of  two  oppo 
site  angles,  as  BAG  and  BDC,  is  measured  by  one-half  the  circum 
ference,  which  is  the  measure  of  two  right  angles,  (54)  and  (I.  19). 


PKOPOSITION  XXII.— THEOREM. 

62.  An  angle  formed  by  a  tangent  and  a  chord  is  measured  by  one- 
half  the  intercepted  arc. 

» 

Let  the  angle  BAG  be  formed  by  the 
tangent  AB  and  the  chord  A  C',  then,  it  is 
measured  by  one-half  the  intercepted  arc 
AMC. 

For,  draw  the  diameter  AD.  The  angle 
BAD  being  a  right  angle  (27),  is  measured 
by  one-half  the  semi-circumference  AMD ; 

and  the  angle  CAD  is  measured  by  one-half  the  arc  CD ;  therefore, 
the  angle  BA  C,  which  is  the  difference  of  the  angles  BAD  and  CAD, 
is  measured  by  one-half  the  difference  of  AMD  and  CD,  that  is, 
by  one-half  the  arc  AMC. 

Also,  the  angle  BrAC  is  measured  by  one-half  the  intercepted  arc 
ANC.  For,  it  is  the  sum  of  the  right  angle  B'AD  and  the  angle 
CAD,  and  is  measured  by  one-half  the  sum  of  the  semi-circumference 
AND  and  the  arc  CD ;  that  is,  by  one-half  the  arc  ANC. 

63.  Scholium.  This  proposition  may  be  treated  as  a  particular  case 
of  Prop.  XXI.  by  an  application  of  the  principle  of  (28).     For,  con 
sider  the  angle  CAD  which  is  measured  by  one-half  the  arc  CD. 
Let  the  side  AC  remain  fixed,  while  the  side  AD,  regarded  as  a 
secant,  revolves  about  A  until  it  arrives  at  the  position  of  the  tangent 


BOOK     II. 


75 


AB'.  The  point  D  will  move  along  the  circumference,  and  will 
ultimately  coincide  with  A,  when  the  line  AD  has  become  a  tangent 
and  the  intercepted  arc  has  become  the  arc  CNA. 


PKOPOSITION  XXIII.— THEOREM. 

64.  An  angle  formed  by  two  chords,  intersecting  within  the  circum 
ference,  is  measured  by  one-half  the  sum  of  the  arcs  intercepted  between 
its  sides  and  between  the  sides  of  its  vertical  angle. 

Let  the  angle  AEC  be  formed  by  the  chords 
AB,  CD,  intersecting  within  the  circumference ; 
then  will  it  be  measured  by  one-half  the  sum 
of  the  arcs  A  C  and  BD,  intercepted  between 
the  sides  of  AEC  and  the  sides  of  its  vertical 
angle  BED. 

For,  join  AD.  The  angle  AEC  is  equal  to  the  sum  of  the  angles 
EDA  and  EAD  (I.  69),  and  these  angles  are  measured  by  one-half 
of  ylOand  one-half  of  BD,  respectively;  therefore,  the  angle  AEC 
is  measured  by  one-half  the  sum  of  the  arcs  AC  and  BD. 


PROPOSITION  XXIV.— THEOREM. 

65.  An  angle  formed  by  two  secants,  intersecting  without  the  circum 
ference,  is  measured  by  one-half  the  difference  of  the  intercepted  arcs. 

Let  the  angle  BA  C  be  formed  by  the  secants  A 

AB  and  AC]  then,  will  it  be  measured  by  one- 
half  the  difference  of  the  arcs  BC  and  DE. 

For,  join  CD.  The  angle  BDC  is  equal  to  the 
sum  of  the  angles  DAC  and  ACD  (I.  69) ;  there 
fore,  the  angle  A  is  equal  to  the  difference  of  the 
angles  BDC  and  A  CD.  But  these  angles  are  meas 
ured  by  one-half  of  BC  and  one-half  of  DE  re 
spectively  ;  hence,  the  angle  A  is  measured  by  one-half  the  differ 
ence  of  BC  and  DE. 


76 


GEOMETRY. 


66.  Corollary.  The  angle  BAE,  formed  by 
a  tangent  AB  and  a  secant  AE,  is  measured 
by  one-half  the  difference  of  the  intercepted 
arcs  BE  and  BC.  For,  the  tangent  AB 
may  be  regarded  as  a  secant  whose  two 
points  of  intersection  are  coincident  at  B 
(28). 

For,  the  same  reason,  the  angle  BAD, 
formed  by  two  tangents  AB  and  AD,  is 

measured  by  one-half  the  difference  of  the  intercepted  arcs  BCD 
and  BED. 

A  proof  may  be  given,  without  using  the  principle  of  (28),  by 
drawing  EB  and  BC. 


PKOBLEMS  OF  CONSTRUCTION. 

Heretofore,  our  figures  have  been  assumed  to  be  constructed  under 
certain  conditions,  although  methods  of  constructing  them  have  not 
been  given.  Indeed,  the  precise  construction  of  the  figures  was  not 
necessary,  inasmuch  as  they  were  only  required  as  aids  in  following 
the  demonstration  of  principles.  We  now  proceed,  first,  to  apply 
these  principles  in  the  solution  of  the  simple  problems  necessary  for 
the  construction  of  the  plane  figures  already  treated  of,  and  then  to 
apply  these  simple  problems  in  the  solution  of  more  complex  ones. 

All  the  constructions  of  elementary  geometry  are  effected  solely 
by  the  straight  line  and  the  circumference,  these  being  the  only  lines 
treated  of  in  the  elements ;  and  these  lines  are  practically  drawn, 
or  described,  by  the  aid  of  the  ruler  and  compasses;  with  the  use  of 
which  the  student  is  supposed  to  be  familiar. 


PEOPOSITION  XXV.— PKOBLEM. 

67.   To  bisect  a  given  straight  line. 

Let  AB  be  the  given  straight  line. 

With  the  points  A  and  B  as  centres,  and  with  a 
radius  greater  than  the  half  of  AB,  describe  arcs 
intersecting  in  the  two  points  D  and  E.  Through 
these  points  draw  the  straight  line  DE,  which  bi 
sects  AB  at  the  point  C.  For,  D  and  E  being 


-i/; 


BOOK     II.  77 

equally  distant  from  A  and  B,  the  straight  line  DE  is  perpendicular 
to  AB  at  its  middle  point  (I.  41). 


PEOPOSITION  XXVI.— PKOBLEM. 

68.  At  a  given  point  in  a  given  straight  line,  to  erect  a  perpendicular 
to  that  line. 

Let  AB  be  the  given  line  and  C  the  given 
point. 

Take  two  points,  D  and  E,  in  the  line  and  at 
equal  distances  from  C.     With  D  and  E  as  cen- 


tres  and  a  radius  greater  than  DC  or  CE  de 
scribe  two  arcs  intersecting  in  F.     Then  CF  is  the  required  perpen 
dicular  (I.  41). 

69.  Another  solution.  Take  any  point   0, 
without  the  given  line,  as  a  centre,  and  with  0  ^ 

a  radius  equal  to  the  distance  from  0  to  C  \    ,.-'"* 

describe  a  circumference  intersecting  A B  in  C      A    D\ 
and  in  a  second  point  D.     Draw  the  diameter 
D  OE,  and  join  EC.     Then  EC  will  be  the  re 
quired  perpendicular :  for  the  angle  ECD,  inscribed  in  a  semicircle, 
is  a  right  angle  (59). 

This  construction  is  often  preferable  to  the  preceding,  especially 
when  the  given  point  C  is  at,  or  near,  one  extremity  of  the  given 
line,  and  it  is  not  convenient  to  produce  the  line  through  that 
extremity.  The  point  0  must  evidently  be  so  chosen  as  not  to  lie  in 
the  required  perpendicular. 


PEOPOSITION  XXVIL— PKOBLEM. 

70.  From  a  given  point  without  a  given  straight  line,  to  let  fall  a  per 
pendicular  to  that  line. 

Let  AB  be  the  given  line  and  C  the  given 
point. 

With  C  as  a  centre,  and  with  a  radius  suf 
ficiently  great,  describe  an  arc  intersecting      A      ID''"- 
AB  in  D  and  E.     With  D  and  E  as  centres 

and  a  radius  greater  than  the  half  of  DE, 

7* 


78 


GEOMETRY. 


describe  two  arcs  intersecting  in  F.  The  line  CF  is  the  required 
perpendicular  (I.  41). 

71.  Another  solution.  With  any  point  0  in 
the  line  AB  as  a  centre,  and  with  the  radius  ...-•*" 

OC,  describe  an  arc  CDE  intersecting  AB      A       o 
in   D.     With  D  as  a  centre  and  a  radius 

**  E 

equal   to  the  distance  DC  describe  an  arc 

intersecting  the  arc  CDE  in  E.  The  line  CE  is  the  required  perpen 
dicular.  For,  the  point  D  is  the  middle  of  the  arc  CDE,  and  the 
radius  OD  drawn  to  this  point  is  perpendicular  to  the  chord 


PROPOSITION  XXVIII.— PROBLEM. 
72.   To  bisect  a  given  arc  or  a  given  angle. 

1st.  Let  AB  be  a  given  arc. 

Bisect  its  chord  AB  by  a  perpendicular  as  in  (67). 
This  perpendicular  also  bisects  the  arc  (16). 


2d.  Let  BA  C  be  a  given  angle.  With  A  as 
a  centre  and  with  any  radius,  describe  an  arc 
intersecting  the  sides  of  the  angle  in  D  and  E. 
With  D  and  E  as  centres,  and  with  equal  radii, 
describe  arcs  intersecting  in  F.  The  straight 
line  AF  bisects  the  arc  DE,  and  consequently 
also  the  angle  BAG  (12). 

73.  Scholium.  By  the  same  construction  each  of  the  halves  of  an 
arc,  or  an  angle,  may  be  bisected  ;  and  thus,  by  successive  bisections, 
an  arc,  or  an  angle,  may  be  divided  into  4,  8,  16,  32,  etc.,  equal 
parts. 


BOOK    II. 


79 


PEOPOSITION  XXIX.— PEOBLEM. 

74.  At  a  given  point  in  a  given  straight  line,  to  construct  an  angle 
equal  to  a  given  angle. 

Let  A  be  the  given  point  in  the  straight  line 
AB,  and  0  the  given  angle. 

With  0  as  a  centre  and  with  any  radius  describe     ( 
an  arc  MN  terminated  by  the  sides  of  the  angle. 
With  A  as  a  centre  and  with  the  same  radius, 
OM,  describe  an  indefinite  arc  BC.     With  B  as  a 
centre  and  with  a  radius  equal  to  the  chord  of 
MN  describe  an  arc  intersecting  the  indefinite  arc     A  ~~          ~~B 
BC  in  D.     Join  AD.     Then  the  angle  BAD  is 
equal  to  the  angle  0.     For  the  chords  of  the  arcs  MN  and  BD  are 
equal ;  therefore,  these  arcs  are  equal  (12),  and  consequently  also  the 
angles  0  and  A  (10). 


PKOPOSITION  XXX.— PKOBLEM. 

75.  Through  a  given  point,  to  draw  a  parallel  to  a  given  straight 
line. 

Let  A  be  the  given  point,  and  B  C  the  given  line. 

From  any  point  B  in  BC  draw  the  straight  f 

line  BAD  through  A.     At  the  point  A,  by  /. 

the  preceding  problem,  construct  the  angle  f  /    ': 

DAE  equal  to  the  angle  ABC.     Then  AE  is  A 

parallel  to  B  C  (I.  55).  -g— i c 

76.  Scholium.  This  problem  is,  in  practice,  more  accurately  solved 
by  the  aid  of  a  triangle,  constructed  of 

wood  or  metal.  This  triangle  has  one 
right  angle,  and  its  acute  angles  are 
usually  made  equal  to  30°  and  60°. 

Let  A  be  the  given  point,  and  BC 
the  given  line.  Place  the  triangle, 
EFD,  with  one  of  its  sides  in  coinci 
dence  with  the  given  line  BC.  Then 
place  the  straight  edge  of  a  ruler  MN 


80 


GEOMETRY. 


against  the  side  EF  of  the  triangle.  Now,  keeping  the  ruler  firmly 
fixed,  slide  the  triangle  along  its  edge  until  the  side  ED  passes 
through  the  given  point  A.  Trace  the  line  EAD  along  the  edge 
ED  of  the  triangle ;  then,  it  is  evident  that  this  line  will  be  parallel 
to  EC. 

One  angle  of  the  triangle  being  made  very  precisely  equal  to  a 
right  angle,  this  instrument  is  also  used  in  practice  to  construct  per 
pendiculars,  with  more  facility  than  by  the  methods  of  (68)  and  (70). 


PKOPOSITION  XXXI.— PKOBLEM. 

77.   Two  angles  of  a  triangle  being  given,  to  find  the  third. 


Let  A  and  B  be  the  given  angles. 

Draw  the  indefinite  line  QM.  From  any 
point  0  in  this  line,  draw  ON  making  the 
angle  MON  =  A,  and  the  line  OP  making 
the  angle  NOP  =  B.  Then  POQ  is  the 
required  third  angle  of  the  triangle  (I.  72). 


PROPOSITION  XXXII.— PROBLEM. 

78.  Two  sides  of  a  triangle  and  their  included  angle  being  given,  to 
construct  the  triangle. 

Let  b  and  c  be  the  given  sides  and  A  their          /          b 

included  angle.  ^- c 

Draw  an  indefinite  line  AE,  and  construct 
the  angle  EAF  =A.  OnAE  take  AC=b, 
and  on  AF  take  AB  =  c;  join  BC.  Then 
ABC  is  the  triangle  required;  for  it  is 
formed  with  the  data. 

With  the  data,  two  sides  and  the  included  angle,  only  one  triangle 
can  be  constructed ;  that  is,  all  triangles  constructed  with  these  data 
are  equal,  and  thus  only  repetitions  of  the  same  triangle  (I.  76). 

79.  Scholium.  It  is  evident  that  one  triangle  is  always  possible, 
whatever  may  be  the  magnitude  of  the  proposed  sides  and  their  in 
cluded  angle. 


BOOK    II.  81 


PROPOSITION  XXXIII.— PROBLEM. 

80.  One  side  and  two  angles  of  a  triangle  being  given,  to  construct 
the  triangle. 

Two  angles  of  the  triangle  being  given, 
the  third  angle  can  be  found  by  (77) ;  and 
we  shall  therefore  always  have  given  the 
two  angles  adjacent  to  the  given  side.  Let, 
then,  c  be  the  given  side,  A  and  B  the  angles 
adjacent  to  it. 

Draw  a  line  AB  =  c;  at  A   make  an 

angle  BAD  =  A,  and  at  B  an  angle  ABE  =  B.     The  lines  AD 
and  BE  intersecting  in  C,  we  have  ABC  as  the  required  triangle. 

With  these  data,  but  one  triangle  can  be  constructed  (I.  78). 

81.  Scholium.  If  the  two  given  angles  are  together  equal  to  or 
greater  than  two  right  angles,  the  problem  is  impossible;  that  is,  no 
triangle  can  be  constructed  with  the  data ;  for  the  lines  AD  and  BC 
will  not  intersect  on  that  side  of  AB  on  which  the  angles  have  been 
constructed. 


PROPOSITION  XXXIV.— PROBLEM. 

82.  The   three   sides   of  a   triangle   being   given,   to   construct   the 
triangle. 

Let  a,  b  and  c  be  the  three  given  sides.  *•  a 

Draw  BC  =  a;  with  C  as  a  centre  and  a 
radius  equal  to  b  describe  an  arc ;  with  B  as 
a  centre  and  a  radius  equal  to  c  describe  a 
second  arc  intersecting  the  first  in  A.  Then, 
ABC  is  the  required  triangle. 

With  these  data  but  one  triangle  can  be  con 
structed  (I.  80). 

83.  Scholium.  The  problem  is  impossible  when  one  of  the  given 
sides  is  equal  to  or  greater  than  the  sum  of  the  other  two  (I.  G6), 

F 


82 


GEOMETRY. 


C"    E 


PKOPOSITION  XXXV.— PEOBLEM. 

84.  Two  sides  of  a  triangle  and  the  angle  opposite  to  one  of  them 
being  given,  to  construct  the  triangle. 

We  shall  consider  two  cases.  / 

1st.  When  the  given  angle  A  is  acute, 
and  the  given  side  a,  opposite  to  it  in  the 
triangle,  is  less  than  the  other  given  side  c. 

Construct  an  angle  DAE  =  A.  In 
one  of  its  sides,  as  AD,  take  AB  =  c ; 
with  B  as  a  centre  and  a  radius  equal  to 
a,  describe  an  arc  which  (since  a  <  c)  will 

intersect  AE  in  two  points,  C'  and  C",  on  the  same  side  of  A.  Join 
BC'  and  BC".  Then,  either  ABC'  or  ABC"  is  the  required  tri 
angle,  since  each  is  formed  with  the  data ;  and  the  problem  has  two 
solutions. 

There  will,  however,  be  but  one  solution,  even  with  these  data,  when 
the  side  a  is  so  much  less  than  the  side  c  as  to  be  just  equal  to  the 
perpendicular  from  B  upon  AE.  For  then  the  arc  described  from  B 
as  a  centre  and  with  the  radius  a,  will  touch  AE  in  a  single  point 
(7,  and  the  required  triangle  will  be  ABCt  right  angled  at  C. 

2d.  When  the  given  angle  A  is  either 
acute,  right  or  obtuse,  and  the  side  a 
opposite  to  it  is  greater  than  the  other 
given  side  c. 

The  same  construction  being  made 
as  in  the  first  case,  the  arc  described 
with  B  as  a  centre  and  with  a  radius  c'\  ^  >'° 

equal  to  a,  will  intersect  AE  in  only  one 

point,  <7,  on  the  same  side  of  A.  Then  ABC  will  be  the  triangle 
required,  and  will  be  the  only  possible  triangle  with  the  data. 

The  second  point  of  intersection,  C',  will  fall  in  EA  produced,  and 
the  triangle  ABC'  thus  formed  will  not  contain  the  given  angle. 

85.  Scholium.  The  problem  is  impossible  when  the  given  angle  A 
is  acute  and  the  proposed  side  opposite  to  it  is  less  than  the  perpen 
dicular  from  B  upon  AE;  for  then  the  arc  described  from  B  will  not 
intersect  AE. 

The  problem  is  also  impossible  when  the  given  angle  is  right,  or 


BOOK     II.  83 

obtuse,  if  the  given  side  opposite  to  the  angle  is  less  than  the  other 
given  side ;  for  either  the  arc  described  from  B  would  not  intersect 
AE,  or  it  would  intersect  it  only  when  produced  through  A.  More 
over,  a  right  or  obtuse  angle  is  the  greatest  angle  of  a  triangle  (I.  70), 
and  the  side  opposite  to  it  must  be  the  greatest  side  (I.  92). 


PKOPOSITION  XXXVI.— PROBLEM. 

86.   The  adjacent  sides  of  a  parallelogram  and  their  included  angle 
being  given,  to  construct  the  parallelogram.  / 

Construct   an   angle   A   equal   to   the   given          si— -.}/> 

angle,  and  take  A  C  and  AB  respectively  equal 
to  the  given  sides.     With  B  as  a  centre  and  a 

•**•  \* 

radius  equal  to  A  C,  describe  an  arc ;  with  C  as 
a  centre  and  a  radius  equal  to  AB,  describe  another  arc,  intersect 
ing  the  first  in  D.     Draw  BD  and  CD.     Then  ABDC  is  a  parallelo 
gram  (I.  107),  and  it  is  the  one  required,  since  it  is  formed  with  the 
data. 

Or  thus:  through  B  draw  BD  parallel  to  AC,  and  through  C 
draw  CD  parallel  to  AB. 


PKOPOSITION  XXXVII.— PKOBLEM. 

87.  To  find  the  centre  of  a  given  circumference,  or  of  a  given  arc. 

Take  any  three  points,  A,  B  and  C,  in  the 
given  circumference  or  arc.  Bisect  the  arcs 
AB,  BC,  by  perpendiculars  to  the  chords  AB, 
J5(7(72);  these  perpendiculars  intersect  in  the 
required  centre  (16). 

88.  Scholium.  The  same  construction  serves  to  describe  a  circum 
ference  which  shall  pass  through  three  given  points  A,  B,  C;  or  to 
circumscribe  a  circle  about  a  given  triangle  ABC,  that  is,  to  describe 
a  circumference  in  which  the  given  triangle  shall  be  inscribed  (56). 


84 


GEOMETRY. 


PROPOSITION  XXXVIII.— PROBLEM. 

89.  At  a  given  point  in  a  given  circumference,  to  draw  a  tangent  to 
the  circumference. 

Let  A  be  the  given  point  in  the  given  circum 
ference.  Draw  the  radius  OA,  and  at  A  draw 
BAG  perpendicular  to  OA;  EC  will  be  the  re 
quired  tangent  (26). 

If  the  centre  of  the  circumference  is  not 
given,  it  may  first  be  found  by  the  preceding 
problem,  or  we  may  proceed  more  directly  as 
follows.  Take  two  points  D  and  E  equidistant 
from  A ;  draw  the  chord  DE,  and  through  A 
draw  BAG  parallel  to  DE.  Since  A  is  the 
middle  point  of  the  arc  DE,  the  radius  drawn 
to  A  will  be  perpendicular  to  DE  (16),  and  con 
sequently  also  to  BC;  therefore  BG  is  a  tangent 
at  A. 

PROPOSITION  XXXIX.— PROBLEM. 

90.  Through  a  given  point  without  a  given  circle  to  draw  a  tangent 
to  the  circle. 

Let  0  be  the  centre  of  the  given  circle  and  P 
the  given  point. 

Upon  OP,  as  a  diameter,  describe  a  circumfer 
ence  intersecting  the  circumference  of  the  given 
circle  in  two  points,  A  and  A.  Draw  PA  and 
PA,  both  of  which  will  be  tangent  to  the  given 
circle.  For,  drawing  the  radii  OA  and  OA,  the 
angles  OAP  and  OA'P  are  right  angles  (59) ; 
therefore  PA  and  PA'  are  tangents  (26). 

In  practice,  this  problem  is  accurately  solved  by  placing  the 
straight  edge  of  a  ruler  through  the  given  point  and  tangent  to  the 
given  circumference,  and  then  tracing  the  tangent  by  the  straight 
edge.  The  precise  point  of  tangency  is  then  determiiled  by  drawing 
a  perpendicular  to  the  tangent  from  the  centre. 

91.  Scholium.  This  problem  always  admits  of  two  solutions.    More 
over,  the  portions  of  the  two  tangents  intercepted  between  the  given 


BOOK     II. 


85 


point  and  the  points  of  tangency  are  equal,  for  the  right  triangles 
POA  and  POA'  are  equal  (L.  83)  ;  therefore,  PA  =  PA. 


PKOPOSITION  XL— PKOBLEM. 

92.   To  draw  a  common  tangent  to  two  given  circles. 
Let  0  and  0'  be  the  centres  of  the  given  circles,  and  let  the 
radius  of  the  first  be  the  greater. 

1st.  To  draw  an  exterior  common  tangent.  With  the  centre  0, 
and  a  radius  OM,  equal  to  the 
difference  of  the  given  radii, 
describe  a  circumference;  and 
from  0'  draw  a  tangent  O'M 
to  this  circumference  (90). 
Join  OM,  and  produce  it  to 
meet  the  given  circumference 
in  A.  Draw  O'A  parallel 'to 

OA,  and  join  AA.  Then  AA  is  a  common  tangent  to  the  two 
given  circles.  For,  by  the  construction,  OM  =  OA  —  O'A',  and 
also  OM=  OA—MA,  whence  MA  =  O'A',  and  AMO'A  is  a  par 
allelogram  (I.  108).  But  the  angle  M  is  a  right  angle ;  therefore, 
this  parallelogram  is  a  rectangle,  and  the  angles  at  A  and  A  are 
right  angles.  Hence,  A  A  is  a  tangent  to  both  circles. 

Since  two  tangents  can  be  drawn  from  0'  to  the  circle  OM,  there 
are  two  exterior  common  tangents  to  the  given  circles,  namely,  AA 
and  BB',  which  meet  in  a  point  T  in  the  line  of  centres  00' 
produced. 

2d.  To  draw  an  interior  common  tangent.     With  the  centre  0 
and  a  radius  OM  equal  to  the  sum  of  the  given  radii,  describe  a  cir 
cumference,  and  from  0'  draw  a  tangent  O'M  to  this  circumference. 
Join  OM,  intersecting  the  given  cir 
cumference  in  A.     Draw  O'A'  par 
allel   to   OA.     Then,  since  OM  = 
OA  +  O'A,  we  have  AM  =  O'A, 
and  AMO'A  is  a  rectangle.    There 
fore,  AA  is  a  tangent  to  both  the 
given  circles. 

There  are  two  interior  common 

8 


86 


GEOMETRY. 


tangents,  A  A'  and  BB',  which  intersect  in  a  point  T  in  the  line  of 
centres,  between  the  two  circles. 

93.  /Scholium.  If  the  given  circles  intersect  each  other,  only  the 
exterior  tangents  are  possible.     If  they  are  tangent  to  each  other 
externally,  the  two  interior  common  tangents  reduce  to  a  single  com 
mon  tangent.     If  they  are  tangent  internally,  the  two  exterior  tan 
gents  reduce  to  a  single  common  tangent,  and  the  interior  tangents 
are  not  possible.     If  one  circle  is  wholly  within  the  other,  there  is 
no  solution. 

PROPOSITION  XLL— PROBLEM. 

94.  To  inscribe  a  circle  in  a  given  triangle. 

Let  ABC  be  the  given  triangle.     Bisect  any  two  of  its  angles,  aa 
B  and  C,  by  straight  lines  meeting  in  0.     From  the  point  0  let  fall 
perpendiculars   OD,   OE,   OF,  upon  the   three 
sides  of  the  triangle ;  these  perpendiculars  will 
be  equal  to  each  other  (I.  129).     Hence,  the 
circumference  of  a   circle,  described  with  the 
centre  0,  and  a  radius  =  OD,  will  pass  through 
the  three  points  D,  E,  F,  will  be  tangent  to  the 
three  sides  of  the  triangle  at  these  points  (26), 
and  will  therefore  be  inscribed  in  the  triangle. 

95.  Scholium.  If  the  sides  of  the  triangle  are  produced  and  the 
exterior  angles  are  bisected,  the  intersections  #',  0",  0'",  of  the 


O" 


BOOK    II. 


87 


bisecting  lines,  will  be  the  centres  of  three  circles,  each  of  which 
will  touch  one  side  of  the  triangle  and  the  two  other  sides  produced. 
In  general,  therefore,  four  circles  can  be  drawn  tangent  to  three  inter 
secting  straight  lines.  The  three  circles  which  lie  without  the  triangle 
have  been  named  escribed  circles. 


PROPOSITION  XLIL— PEOBLEM. 

96.  Upon  a  given  straight  line,  to  describe  a  segment  which  shall 
contain  a  given  angle. 

Let  AB  be  the  given  line.     At  the  point  B  construct  the  angle 
ABC  equal  to  the  given  angle.     Draw  BO  per 
pendicular  to  BC,  and  DO  perpendicular  to 
AB  at  its  middle  point  D,  intersecting  B  0  in  0. 
With  0  as  a  centre,  and  radius  OB  describe  the 
circumference  AMBN.     The  segment  AMB  is 
the  required  segment.     For,  the  line  BC,  being 
perpendicular  to  the  radius  OB,  is  a  tangent  to 
the  circle;  therefore,  the  angle  ABC  is  meas 
ured  by  one-half  the  arc  ANB  (62),  which  is  also  the  measure  of 
any  angle  AMB  inscribed  in  the  segment  AMB  (57).     Therefore, 
any  angle  inscribed  in  this  segment  is  equal  to  the  given  angle. 

97.  Scholium.  If  any  point  P  is  taken  within  the  segment  AMB, 
the  angle  APB  is  greater  than  the  inscribed  angle 

AMB  (I.  74) ;  and  if  any  point  Q  is  taken  without 
this  segment,  but  on  the  same  side  of  the  chord  AB 
as  the  segment,  the  angle  A  QB  is  less  than  the  in 
scribed  angle  AMB.  Therefore,  the  angles  whose 
vertices  lie  in  the  arc  AMB  are  the  only  angles  of 
the  given  magnitude  whose  sides  pass  through  the 
two  points  A  and  B ;  hence,  the  arc  AMB  is  the 
locus  of  the  vertices  of  all  the  angles  of  the  given 
magnitude  whose  sides  pass  through  A  and  B. 

If  any  point  M'  be  taken  in  the  arc  AM'B,  the  angle  AMB  is  the 
supplement  of  the  angle  AM'B  (61)  ;  and  if  BM'  be  produced  to 
B',  the  angle  AM' B'  is  also  the  supplement  of  AM'B;  therefore 
AM'B'  =  AMB.  Hence  the  vertices  of  all  the  angles  of  the  given 
magnitude  whose  sides,  or  sides  produced,  pass  through  A  and  B,  lie 


00  GEOMETRY. 

in  the  circumference  AMBM' ;  that  is,  the  locus  of  the  vertices  of  all 
the  angles  of  a  given  magnitude  whose  sides,  or  sides  produced,  pass 
through  two  fixed  points,  is  a  circumference  passing  through  these  points, 
and  this  locus  may  be  constructed  by  the  preceding  problem. 

It  may  here  be  remarked,  that  in  order  to  establish  a  certain  line 
as  a  locus  of  points  subject  to  certain  given  conditions,  it  is  necessary 
not  only  to  show  that  every  point  in  that  line  satisfies  the  conditions, 
but  also  that  no  other  points  satisfy  them ;  for  the  asserted  locus 
must  be  the  assemblage  of  all  the  points  satisfying  the  given  condi 
tions  (I.  40). 


INSCRIBED  AND  CIRCUMSCRIBED  QUADRILATERALS. 

98.  Definition.  An  inscriptible  quadrilateral  is  one  which  can  be 
inscribed  in  a  circle ;  that  is,  a  circumference  can  be  described  pass 
ing  through  its  four  vertices. 


PROPOSITION  XLIIL— THEOREM. 

99.  A  quadrilateral  is  inscriptible  if  two  opposite  angles  in  it  are 
supplements  of  each  other. 

Let  the  angles  A  and  C,  of  the  quadrilateral 
ABCD,  be  supplements  of  each  other.  De 
scribe  a  circumference  passing  through  the 
three  vertices  B,  C,  D;  and  draw  the  chord 
BD.  The  angle  A,  being  the  supplement  of 
C,  is  equal  to  any  angle  inscribed  in  the  seg 
ment  BMD  (61)  ;  therefore  the  vertex  A  must 
be  on  the  arc  AMD  (97),  and  the  quadrilateral  is  inscribed  in  the 
circle. 

100.  Scholium.  This  proposition  is  the  converse  of  (61). 


BOOK    II. 


89 


PKOPOSITION  XLIV.— THEOEEM. 

101.  In  any  circumscribed  quadrilateral,  the  sum  of  two  opposite  sides 
is  equal  to  the  sum  of  the  other  two  opposite  sides. 

Let  AB  CD  be  circumscribed  about  a  circle ; 
then, 

AB  +  DC  =  AD-}-  BC. 

For,  let  E,  F,  G,  H,  be  the  points  of  contact 
of  the  sides ;  then  we  have  (91), 

AE=AH,    BE=BF,     CG  =  CF, 

Adding  the  corresponding  members  of  these  equalities,  we  have 
AE  +  BE+  CG  +  DG  =  AH+DH+BF  +  CF, 

that  is, 

AB      DC=AD  +  BC. 


PKOPOSITION  XLV.— THEOKEM. 

102.  Conversely,  if  the  sum  of  two  opposite  sides  of  a  quadrilateral 
is  equal  to  the  sum  of  the  other  two  sides,  the  quadrilateral  may  be  cir 
cumscribed  about  a  circle. 

In  the  quadrilateral  ABCD,  letAB  +  DC  = 
AD  -f  BC;  then,  the  quadrilateral  can  be  cir 
cumscribed  about  a  circle. 

Since  the  sum  of  the  Tour  angles  of  the  quad 
rilateral  is  equal  to  four  right  angles,  there  must 
be  two  consecutive  angles  in  it  whose  sum-  is  not 
greater  than  two  right  angles ;  let  B  and  C  be 
these  angles.     Let  a  circle  be  described  tangent  to  the  three  sides 
AB,  BC,  CD,  the  centre  of  this  circle  being  the  intersection  of  the 
bisectors  of  the  angles  B  and  C;  then  it  is  to  be  proved  that  this 
circle  is  tangent  also  to  the  fourth  side  AD. 

From  the  point  A  two  tangents  can  be  drawn  to  the  circle  (90). 
One  of  these  tangents  being  AB,  the  other  must  be  a  line  cutting 
CD  (or  CD  produced) ;  for,  the  sum  of  the  angles  B  and  C  being 
not  greater  than  two  right  angles,  it  is  evident  that  no  straight  line 


90  GEOMETRY. 

can  be  drawn  from  A,  falling  on  the  same  side  of  BA  with  CD,  and 

not  cutting  the  circle,  which  shall  not  cut  CD. 

This  second  tangent,  then,  must  be  either  AD 

or  some  other  line,  AM,  cutting  CD  in  a  point  M 

differing  from  D.     If  now  AM  is  a  tangent, 

AB  CM  is  a  circumscribed  quadrilateral,  and  by 

the  preceding  proposition  we  shall  have 

AB+  CM=AM  +  BC. 
But  we  also  have,  by  the  hypothesis  of  the  present  proposition, 

AB  +  DC=AD  +  BC. 

Taking  the  difference  of  these  equalities,  we  have 
DM  =  AM— AD-, 

that  is,  one  side  of  a  triangle  is  equal  to  the  difference  of  the  other  two, 
which  is  absurd.  Therefore,  the  hypothesis  that  the  tangent  drawn 
from  A  and  cutting  the  line  CD,  cuts  it  in  any  other  point  than  D, 
leads  to  an  absurdity ;  therefore,  that  hypothesis  must  be  false,  and 
the  tangent  in  question  must  cut  CD  in  D,  and  consequently  coincide 
with  AD.  Hence,  a  circle  has  been  described  which  is  tangent  to 
the  four  sides  of  the  quadrilateral ;  and  the  quadrilateral  is  circum 
scribed  about  the  circle. 

103.  Scholium.  The  method  of  demonstration  employed  above  is 
called  the  indirect  method,  or  the  reductio  ad  absurdum.  At  the 
outset  of  a  demonstration,  or  at  any  stage  of  its  progress,  two  or 
more  hypotheses  respecting  the  quantities  under  consideration  may 
be  admissible  so  far  as  has  been  proved  up  to  that  point.  If,  now, 
these  hypotheses  are  such  that  one  must  be  true,;and  only  one  can 
be  true,  then,  when  all  except  one  are  shown  to  be  absurd,  that  one 
must  stand  as  the  truth. 

While  admitting  the  validity  of  this  method,  geometers  usually 
prefer  the  direct  method  whenever  it  is  applicable.  There  are,  how 
ever,  propositions,  such  as  the  preceding,  of  which  no  direct  proof  is 
known,  or  at  least  no  proof  sufficiently  simple  to  be  admitted  into 
elementary  geometry.  We  have  already  employed  the  reductio  ad 
absurdum  in  several  cases  without  presenting  the  argument  in  full ; 
see  (I.  47),  (I.  85),  (27). 


BOOK   III. 

PROPORTIONAL  LINES.    SIMILAR  FIGURES. 
THEORY  OF  PROPORTION. 

1.  DEFINITION.  One  quantity  is  said  to  be  -proportional  to  another 
when  the  ratio  of  any  two  values,  A  and  B,  of  the  first,  is  equal  to 
the  ratio  of  the  two  corresponding  values,  A  and  B',  of  the  second ; 
so  that  the  four  values  form  the  proportion 
A  :  B  =  A  :  B', 

A       A 

or  —  =  — ',' 

This  definition  presupposes  two  quantities,  each  of  which  can  have 
various  values,  so  related  to  each  other  that  each  value  of  one  cor 
responds  to  a  value  of  the  other.  An  example  occurs  in  the  case  of 
an  angle  at  the  centre  of  a  circle  and  its  intercepted  arc.  The 
angle  may  vary,  and  with  it  also  the  arc ;  but  to  each  value  of  the 
angle  there  corresponds  a  certain  value  of  the  arc.  It  has  been 
proved  (II.  51)  that  the  ratio  of  any  two  values  of  the  angle  is  equal 
to  the  ratio  of  the  two  corresponding  values  of  the  arc ;  and  in  ac 
cordance  with  the  definition  just  given,  this  proposition  would  be 
brieflv  expressed  as  follows :  "  The  angle  at  the  centre  of  a  circle  is 
proportional  to  its  intercepted  arc." 

2.  Definition.  One  quantity  is  said  to  be  reciprocally  proportional 
to  another  when  the  ratio  of  two  values,  A  and  B,  of  the  first,  is 
equal  to  the  reciprocal  of  the  ratio  of  the  two  corresponding  values, 
A  and  I?,  of  the  second,  so  that  the  four  values  form  the  proportion 

A  :  B  =  B'  :  A, 
A       Br  A' 


B^T^'-v' 


91 


92  GEOMETRY. 

For  example,  if  the  product  p  of  two  numbers,  x  and  y,  is  given, 

so  that  we  have 

xy=p, 

then,  x  and  y  may  each  have  an  indefinite  number  of  values,  but  as 
x  increases  y  diminishes.  If,  now,  A  and  B  are  two  values  of  x, 
while  A'  and  B'  are  the  two  corresponding  values  of  y,  we  must  have 


BxBf=p, 
whence,  by  dividing  one  of  these  equations  by  the  other, 

^x^-i 

SX  B'" 

and  therefore 

—  =  L  =  iL. 

B  ~  A   '~A'' 
B' 

that  is,  two  numbers  whose  product  is  constant  are  reciprocally  propor 
tional. 

3.  Let  the  quantities  in  each  of  the  couplets  of  the  proportion 

|  =  |1     o*A:B  =  A':B',  [1] 

be  measured  by  a  unit  of  their  own  kind,  and  thus  expressed  by 
numbers  (II.  42)  ;  let  a  and  b  denote  the  numerical  measures  of  A  and 
B,  a'  and  b'  those  of  A'  and  B'  ;  then  (II.  43), 

A  =  a  4L  =  <?L 

B~  b  B'       b' 

and  the  proportion  [1]  may  be  replaced  by  the  numerical  proportion, 

a       ae  ,  ,     ,, 

-  =  —j     or  a  :  b  =  a   :  b  . 
b       b 

4.  Conversely,  if  the  numerical  measures  a,  b,  a',br,  of  four  quan 
tities  A,  B,  A,  B',  are  in  proportion,  these  quantities  themselves  are 
in  proportion,  provided  that  A  and  B  are  quantities  of  the  same  kind, 
and  A  and  B'  are  quantities  of  the  same  kind  (though  not  neces 
sarily  of  the  same  kind  as  A  and  B)  ;  that  is,  if  we  have 

a  :  b  =  a'  :  b', 


BOOK     III.  93 


we  may,  under  these  conditions,  infer  the  proportion 

A  :  B  =  A'  :  B'. 
5.  Let  us  now  consider  the  numerical  proportion 

a  :  b  =  a'  bf. 
Writing  it  in  the  form 


and  multiplying  both  members  of  this  equality  by  W,  we  obtain 

ab'  =  a'b, 

whence  the  theorem  :   the  product  of  the  extremes  of  a  (numerical) 
proportion  is  equal  to  the  product  of  the  means. 

Corollary.  If  the  means  are  equal,  as  in  the  proportion  a  :  b  =  b  :  c, 
we  have  b2  =  ac,  whence  b  =  i/ac  ;  that  is,  a  mean  proportional  be 
tween  two  numbers  is  equal  to  the  square  root  of  their  product. 

6.  Conversely,  if  the  product  of  two  numbers  is  equal  to  the  product 
of  two  others,  either  two  may  be  made  the  extremes,  and  the  other  two  the 
means,  of  a  proportion.  For,  if  we  have  given 

ab'  =  a'b, 
then,  dividing  by  bb',  we  obtain 

a       a'  ,          .    ,, 

-  =  —  »     or  a  :  b  =  a   :  b  . 
b        b' 

Corollary.  The  terms  of  a  proportion  may  be  written  in  any  order 
which  will  make  the  products  of  the  extremes  equal  to  the  product 
of  the  means.  Thus,  any  one  of  the  following  proportions  may  be 
inferred  from  the  given  equality  abf  =  a'b: 

a   :  b  =af  :bf, 

a   :a'  =  b   :b', 

b    :a  =b':a'y 

b    :b'  =  a  :ar, 

b'  :  a'  =  b   :  a,  etc. 

Also,  any  one  of  these  proportions  may  be  inferred  from  any  other. 
7.  Definitions.  When  we  have  given  the  proportion 

a  :  b  =  a'  :  b', 


94  GEOMETRY. 

and  infer  the  proportion 

a:  a'  =  bib', 

the  second  proportion  is  said  to  be  deduced  by  alternation. 
When  we  infer  the  proportion 


this  proportion  is  said  to  be  deduced  by  inversion. 

8.  It  is  important  to  observe,  that  when  we  speak  of  the  products 
of  the  extremes  and  means  of  a  proportion,  it  is  implied  that  at  least 
two  of  the  terms  are  numbers.     If,  for  example,  the  terms  of  the 

proportion 

A:B  =  A:Bf, 

are  all  lines,  no  meaning  can  be  directly  attached  to  the  products 
A  X  B',  B  X  Af,  since  in  a  product  the  multiplier  at  least  must  be 
a  number. 

But  if  we  have  a  proportion  such  as 

A  :  B  =•  m  :  n, 

in  which  m  and  n  are  numbers,  while  A  and  B  are  any  two  quanti 
ties  of  the  same  kind,  then  we  may  infer  the  equality  nA  =  mB. 

Nevertheless,  we  shall  for  the  sake  of  brevity  often  speak  of  the 
product  of  two  lines,  meaning  thereby  the  product  of  the  numbers 
which  represent  those  lines  when  they  are  measured  by  a  common  unit. 

9.  If  A  and  B  are  any  two  quantities  of  the  same  kind,  and  m 
any  number  whole  or  fractional,  we  have,  identically, 

mA       A  s 


that  is,  equimultiples  of  two  quantities  are  in  the  same  ratio  as  the 
quantities  themselves. 

Similarly,  if  we  have  the  proportion 

A  :  B  =  A'  :  B', 

and  if  m  and  n  are  any  two  numbers,  we  can  infer  the  proportions 
mA  :  mB  =  nA  :  nB', 
mA  :  nB  =  mA':  nB'. 


BOOK     III. 


95 


10.   Composition  and  division.  Suppose  we  have  given  the  propor 
tion 

I=F       :  -    w 

in  which  A  and  B  are  any  quantities  of  the  same  kind,  and  A  and 
B'  quantities  of  the  same  kind.  Let  unity  be  added  to  both  mem 
bers  of  [1] ;  then 


or,  reducing, 


A  +  B  =  A  +  B'      \ 
B  B' 


and  dividing  this  by  [1], 

A  +  B      A  +  B' 
A  A 


[2] 


results  which  are  briefly  expressed  by  the  theorem,  if  four  quantities 
are  in  proportion,  they  are  in  proportion  by  composition;  the  term 
composition  being  employed  to  express  the  addition  of  antecedent 
and  consequent  in  each  ratio. 

If  we  had  subtracted  unity  from  both  members  of  [1],  we  should 
have  found 

A  —  B       A'  —  Br 


[3] 


A 


results  which  are  briefly  expressed  by  the  theorem,  if  four  quantities 
are  in  proportion,  they  are  in  proportion  by  division ;  where  the  term 
division  is  employed  to  express  the  subtraction  of  consequent  from 
antecedent  in  each  ratio,  this  subtraction  being  conceived  to  divide, 
or  to  separate,  the  antecedent  into  parts. 
The  quotient  of  [2]  divided  by  [3]  is 

A  4-  B       A  4-  B' 


A  —  B       A  —  B'' 

that  is,  if  four  quantities  are  in  proportion,  they  are  in  proportion  by 
composition  and  division. 


96  GEOMETRY. 

11.  Definition.  A  continued  proportion  is  a  series  of  equal  ratios,  as 

A:B  =  A'.B'  =  A':  B"  =  A"  :  B'"  =--  etc. 

12.  Let  r  denote  the  common  value  of  the  ratio  in  the  continued 
proportion  of  the  preceding  article ;  that  is,  let 

_  A_A^_A^__A^_ 
~  B~  B'~  B"~  B'"~ 

then,  we  have 

A  =  Br,    A  =  Bfr,    A' =  B"  r,    A"  =  B'" r,  etc., 
and  adding  these  equations, 

A  +  A'  +  A'  +  A"  +  etc.  =  (B  +  B'  +  B"  +  B'"  +  etc.)  r, 
whence 

A  +  A  +  A"  +  uiw  +  etc.  =_  r  =  ;4  =  J/  __  etc  . 

#  _j_  £'  -j-  JB"  -f-  £'"  -f  etc.  ^       jB' 

that  is,  the  sum  of  any  number  of  the  antecedents  of  a  continued  pro 
portion  is  to  the  sum  of  the  corresponding  consequents  as  any  antecedent 
is  to  its  consequent. 

If  any  antecedent  and  its  corresponding  consequent  be  taken  with 
the  negative  sign,  the  theorem  still  holds,  provided  we  read  algebraic 
sum  for  sum. 

In  this  theorem  the  quantities  A,  B,  C,  etc.,  must  all  be  quantities 
of  the  same  kind. 

13.  If  we  have  any  number  of  proportions,  as 

a  :  b  =  c  :  d, 
a':b'  =  cf:df, 
a":b"  =  c"  :d",etc.', 
then,  writing  them  in  the  form, 

a_c        a'  __J_       a^  _^_ 
b~~d'       b'~d'        b"-d"* 

and  multiplying  these  equations  together,  we  have 


BOOK     III.  97 


bb'  b"  ...        dd'd"... 
or 

a  a'  a"  .  .  .  :  b  b'  b"  .  .  .  =  c  c'  e"  .  .  .  :  d  d'  d"  .  .  .  ; 

that  is,  if  the  corresponding  terms  of  two  or  more  proportions  are  mul 
tiplied  together,  the  products  are  in  proportion. 

If  the  corresponding  terms  of  the  several  proportions  are  equal, 
that  is,  if  a  =  a'  =  a",  b  =  b'  =  b",  etc.,  then  the  multiplication 
of  two  or  more  proportions  gives 

a2  :  b2  =  c*  :  d2, 
as:b3  =  c3:  d3; 

that  is,  if  four  numbers  are  in  proportion,  like  powers  of  these  numbers 
are  in  proportion. 

14.  If  A,  B  and  C  are  like  quantities  of  any  kind,  and  if 

A  B 

-  =  m>and--  =  n, 

then 


If  A,  B.  and  C  were  numbers,  this  would  be  proved,  arithmetically, 
by  simply  omitting  the  common  factor  B  in  the  multiplication  of  the 
two  fractious  ;  but  when  they  are  not  numbers  we  cannot  regard  B 
as  a  factor,  or  multiplier,  and  therefore  we  should  proceed  more 
strictly  as  follows.  By  the  nature  of  ratio  we  have 

A  =  BXm,    B=  CXn, 
therefore,  putting  C  X  n  for  B,  we  have 

A=CXnXm=CXmn, 
that  is, 

A 

~  =  mn; 

a  result  usually  expressed  as  follows  :  the  ratio  of  the  first  of  three 
quantities  to  the  third  is  compounded  of  the  ratio  of  the  first  to  the  second 
and  the  ratio  of  the  second  to  the  third. 
9  G 


98  GEOMETRY. 

PROPORTIONAL   LINES. 
PROPOSITION  I.—  THEOREM. 

15.  A  parallel  to  the  base  of  a  triangle  divides  the  other  two  sides 
proportionally. 

Let  DE  be  a  parallel  to  the  base,  EG,  of  the  triangle  ABC;  then, 

AB:AD  =  AC:AE. 

1st.  Suppose  the  lines  AB,  AD,  to  have  a 
common  measure  which  is  contained,  for  exam 

ple,  7  times  in  AB,  and  4  times  in  AD  ;  so  that  J  -  \^ 

if  AB  is  divided  into  7  parts  each  equal  to  the  /-  .............. 

common  measure,  AD  will  contain  4  of  these  /  .....  *  ........  ' 

parts.     Then  the  ratio  of  AB  to  AD  is  7  :  4  * 

(11.43);  that  is 

AH 


Through  the  several  points  of  division  of  AB,  draw  parallels  to  the 
base;  then  AC  mil  be  divided  into  7  equal  parts  (I.  125),  of  which 
AE  will  contain  4.  Hence  the  ratio  of  AC  to  AE  is  7  :  4  ;  that  is, 


— 
AE     I 

Therefore,  we  have 

AB  _AC 

AD    .  AE 


or 


AB  i  AV  =  AC  :  AE. 


2d.  If  AB  and  AD  are  incommensurable,  suppose  one  of  tlienv 
as  AD,  to  be  divided  into  any  number  n  of  equal  parts ;  then,  AB 
will  contain  a  certain  number  m  of  the&e  parts  plus  a  remainder  less 
than  one  of  these  parts.  The  numerical  expression  of  the  ratio 

—  will  then  be  ->  correct  within  -  (II.  48)..  Drawing  parallels  to 
AD  n  n 

BC,  through  the  several  points  of  division  of  AB,  the  line  AEvfill 
be  divided  into  n  equal  parts,  and  the  line  AC  will  contain  m  such 
parts  plus  a  remainder  less  than  one  of  the  parts.  Therefore,  the 


BOOK    III.  99 

AC  m  1 

numerical  expression  of  the  ratio  --  will  also  be  -,  correct  within  -• 

AE  n  n 

Since,  then,  the  two  ratios  always  have  the  same  approximate  nu 
merical  expression,  however  small  the  parts  into  which  AD  is  divided, 
these  ratios  must  be  absolutely  equal  (II.  49),  and  we  have,  as  before, 


AD 

or  AB  :  AD  =  AC  :  AE.  [1] 

16.  Corollary  I.  By  division  (10),  the  proportion  [1]  gives 

AB  —  AD:AB  =  AC—  AE  :  AC, 
or  DB:AB  =  EC:AC. 

Also,  if  the  parallel  DE  intersect  the  sides  BA 
and  CA  produced  through  J.,  we  find,  as  in  the 
preceding  demonstration, 

AB:AD  =  AC:AE, 
from  which,  by  composition  (10), 

AB-}-  AD  :AB  =  AC+  AE  :  AC, 
or  DB:AB  =  EC:AC. 

17.  Corollary  II.  By  alternation  (7),  the  preceding  proportions 
give 

AB:AC=AD:AE, 

DB:EC=AB:ACt 

which  may  both  be  expressed  in  one  continued  proportion, 

AB       AD       DB 
AC~  AE~~  EC 

This  proportion  is  indeed  the  most  general  statement  of  the  proposi 
tion  (15),  which  may  also  be  expressed  as  follows  :  if  a  straight  line 
is  drawn  parallel  to  the  base  of  a  triangle,  the  corresponding  segments 
on  the  two  sides  are  in  a  constant  ratio. 


100 


GEOMETRY. 


18.  Corollary  III.  If  two  straight  lines  MN,  M'N',  are  intersected 
by  any  number  of  parallels  A  A',  BBf,  CCf,  etc.,  the  corresponding 
segments  of  the  two  lines  are  proportional. 
For,  let  the  two  lines  meet  in  0  ;  then,  by 
Corollary  II., 

OB_  _=  BC_^OC^=  CD_  ^ 
OB'~~B'Cr~OC'     C'D'^ 


OA'A'Bt 

whence,  by 
AB 


BD 
B'D' 


AB'~~  B'C'       C'D'      AC' 

If  MN  and  M'N'  were  parallel,  this  proportion  would  still  hold, 
since  we  should  then  have  AB  =  AB',  BC  =  B'C',  etc. 


PROPOSITION  II.— THEOREM. 


19.  Conversely,  if  a  straight  line  divides  two  sides  of  a  triangle  pro 
portionally,  it  is  parallel  to  the  third  side. 

Let  DE  divide  the  sides  AB,  A  C,  of  the  triangle 
ABC,  proportionally;  then,  Dj£is  parallel  to  BC. 

For,  if  DE  is  not  parallel  to  BC,  let  some  other 
line  DE',  drawn  through  D,  be  parallel  to  BC. 
Then,  by  the  preceding  theorem, 

AB:AD  =  AC-.AE'. 
But,  by  hypothesis,  we  have 

=  AC:AE, 


whence  it  follows  that  AE'  =  AE,  which  is  impossible  unless  DE' 
coincides  with  DE.     Therefore,  DE  is  parallel  to  BC. 
20.  Scholium.  The  converse  of  (18)  is  not  generally  true. 


PROPOSITION  III.— THEOREM. 

21.  In  any  triangle,  the  bisector  of  an  angle,  or  the  bisector  of  its 
exterior  angle,  divides  the  opposite  side,  internally  or  externally,  into 
segments  which  are  proportional  to  the  adjacent  sides. 


BOOK     III.  101 

1st.  Let  AD  bisect  the  angle  A  of 
the  triangle  ABC;  then, 

DE  :  DC '=  AB  :  AC. 

For,  through  B  draw  BE  parallel  

to  DA,  meeting  CA  produced  in  E. 

The  angle  ABE  =  BAD  (I.  49),  and  the  angle  AEB  =  CAD 
(I.  51) ;  and,  by  hypothesis,  the  angle  BAD  =  CAD ;  therefore,  the 
angle  ABE  =  AEB,  and  AE  =  AB  (I.  90). 

Now,  in  the  triangle  CEB,  AD  being  parallel  to  EB,  we  have  (17), 

DB:DC=  AEiAC, 
or  DB:DC=AB:AC; 

that  is,  the  side  BC  is  divided  by  AD  internally  into  segments  pro 
portional  to  the  adjacent  sides  A  B  and  A  C. 

2d.  Let  AD'  bisect  the  exterior  angle  BAE;  then, 

D'B:D'C  =  AB:AC. 

For,  draw  BE'  parallel  to  D'A;  then,  ABE'  is  an  isosceles  tri 
angle,  and  AE1  =  AB.     In  the  triangle  CAD',  we  have  (17), 

D'B:DfC=AE'i  AC, 
or  D'B:D'C  =  AB  :  AC; 

that  is,  the  side  BC  is  divided  by  AD'  externally  into  segments  pro 
portional  to  the  adjacent  sides  AB  and  AC. 

22.  Scholium.  When  a  point  is  taken  on  a  given  finite  line,  or  on 
the  line  produced,  the  distances  of  the  point  from  the  extremities  of 
the  line  are  called  the  segments,  internal  or  external,  of  the  line. 
The  given  line  is  the  sum  of  two  internal  segments,  or  the  difference 
of  two  external  segments. 

23.  Corollary.  If  a  straight  line,  drawn  from  the  vertex  of  any 
angle  of  a  triangle  to  the  opposite  side,  divides  that  side  internally 
in  the  ratio  of  the  other  two  sides,  it  is  the  bisector  of  the  angle ;  if 
it  divides  the  opposite  side  externally  in  that  ratio,  it  is  the  bisector 
of  the  exterior  angle.     (To  be  proved). 


102 


GEOMETRY. 


SIMILAR  POLYGONS. 

24.  Definitions.  Two  polygons  are  similar,  when  they  are  mutually 
equiangular  and  have  their  homologous  sides  proportional. 

In  similar  polygons,  any  points,  angles  or  lines,  similarly  situated 
in  each,  are  called  homologous. 

The  ratio  of  a  side  of  one  polygon  to  its  homologous  side  in  the 
other  is  called  the  ratio  of  similitude  of  the  polygons. 


PROPOSITION  IV.— THEOREM. 

25.   Two  triangles  are  similar,  when  they  are  mutually  equiangular. 

Let  ABC,  AB'C',\>Q  mutually  equiangular  triangles,  in  which 
A  =  A',  B  =  B',  C  ==€';  then, 
these  triangles  are  similar. 

For,  place  the  angle  A'  upon  its 
equal  angle  A,  and  let  B'  fall  at  6 
and  C'  at  c.  Since  the  angle  Abe  is 
equal  to  B,  be  is  parallel  to  BG 
(I.  55),  and  we  have  (15), 


or 

In  the  same  manner,  it  is  proved  that 

AB:  AB'  =  BC'.B'C'', 
and,  combining  these  proportions, 

AB  _,  AC  =_  EC  m 

AB'  ~~~  AC'~  B'C1' 

Therefore,  the  homologous  sides  are  proportional,  and  the  triangles 
are  similar  (24). 

26.  Corollary.  Two  triangles  are  similar  when  two  angles  of  the 
one  are  respectively  equal  to  two  angles  of  the  other  (I.  73). 

27.  Scholium  I.  The  homologous  sides  lie  opposite  to  equal  angles. 

28.  Scholium  II.  The  ratio  of  similitude  (24)  of  the  two  similar 
triangles,  is  any  one  of  the  equal  ratios  in  the  continued  propor 
tion  [1]. 


BOOK    III. 


103 


29.  Scholium  III.  In  two  similar  triangles,  any  two  homologous 
lines  are  in  the  ratio  of  similitude  of 
the  triangles.  For  example,  the  per 
pendiculars  AD,  A'D',  drawn  from  the 
homologous  vertices  A,  A',  to  the  op 
posite  sides,  are  homologous  lines  of 
the  two  triangles ;  and  the  right  tri 
angles  ABD,  A'B'D',  being  similar 
(25),  we  have 

AD        AB        AC 
r  A'B'  ~ 


BC 


AD' 


AC'       B'C' 


In  like  manner,  if  the  lines  AD,  AD',  were  drawn  from  A,  A',  to 
the  middle  points  of  the  opposite  sides,  or  to  two  points  which  divide 
the  opposite  sides  in  the  same  ratio  in  each  triangle,  these  lines 
would  still  be  to  each  other  in  the  ratio  of  similitude  of  the  two 
triangles. 

PKOPOSITION  V.— THEOEEM. 

30.   Two  triangles  are  similar,  when  their  homologous  sides  are  pro 
portional. 
In  the  triangles  ABC,  A  B'C',  let 

AB        AC        BC 


A'B'       AC'       B'C'' 


[1] 


then,  these  triangles  are  similar. 

For,  on  AB  take  Ab  =  A'B',  and 
draw  be  parallel  to  BC.  Then,  the 
triangles  Abe  and  ABC  are  mutually 
equiangular,  and  we  have  (25), 

AB^AB^  ___AC_2W 

Ab  ° 


A'B' 


Ac       Be 


Comparing  this  with  the  given  proportion  [1],  we  see  that  the  first 
ratio  is  the  same  in  both ;  hence  the  second  and  third  ratios  in  each 
are  equal  respectively,  and,  the  numerators  being  the  same,  the 
denominators  are  equal;  that  is,  AC'  =  Ac,  and  B'C'  =  Be. 
Therefore,  the  triangles  A  B'C'  and  Abe  are  equal  (I.  80)  ;  and  since 
Abe  is  similar  to  ABC,  A'B' C'  is  also  similar  to  ABC. 


104  GEOMETRY. 

31.  Scholium.  In  order  to  establish  the  similarity  of  two  polygons 
according  to  the  definition  (24),  it  is  necessary,  in  general,  to  show 
that  they  fulfill  two  conditions  :   1st,  they  must  be  mutually  equi 
angular,  and  2d,  their  homologous  sides  must  be  proportional.     In 
the  case  of  triangles,  however,  either  of  these  conditions  involves  the 
other  ;  and  to  establish  the  similarity  of  two  triangles  it  will  be  suf 
ficient  to  show,  either  that  they  are  mutually  equiangular,  or  that 
their  homologous  sides  are  proportional. 

PKOPOSITION  VI.—  THEOEEM. 

32.  Two  triangles  are  similar,  when  an  angle  of  the  one  is  equal  to 
an  angle  of  the  other,  and  the  sides  including  these  angles  are  propor- 
portional. 

In  the  triangles  ABC,  A'B'C',  let  A  A> 

A  =  A,  and 

AB         AC  m 
AB'~~  AC'' 

then,  these  triangles  are  similar. 

For,  place  the  angle  A  upon  its 

equal  angle  A;  let  B'  fall  at  b,  and  C'  at  c.     Then,  by  the  hy 
pothesis, 


Ab        Ac' 

Therefore,  be  is  parallel  to  BC  (19),  and  the  triangle  Abe  is  similar 
to  ABC  (25).  But  Abe,  is  equal  to  A'B'C'-,  therefore,  A'B'C'  is 
also  similar  to  ABC. 

PKOPOSITION  VII.—  THEOEEM. 

33.   Two  triangles  are  similar,  when  they  have  their  sides  parallel 
each  to  each,  or  perpendicular  each  to  each. 

Let  ABC,  abc  have  their  sides  par 
allel  each  to  each,  or  perpendicular 
each  to  each  ;  then,  these  triangles  are 
similar. 

For,  when  the  sides  of  two  angles       B 


BOOK      III 


105 


are  parallel  each  to  each,  or  perpen 
dicular  each  to  each,  these  angles  are 
either  equal,  or  supplements  of  each 
other,  (I.  60,  62,  63).  In  the  present 
case,  therefore,  three  hypotheses  may  be 
made,  namely,  denoting  a  right  angle 


1st  hyp.  A  +  a  =  2E,    B  +  b  =  2R,     C  +  c  =  2R-, 
2d     "  A  =  a,       B  +  b  =  2E,     C  +  c  =  2J?; 

3d     "  yl  =  a,  B  =  b,  whence  C  =  c. 

The  1st  and  2d  hypotheses  cannot  be  admitted,  since  the  sum  of  all 
the  angles  of  the  two  triangles  would  then  exceed  four  right  angles 
(I.  68).  The  3d  hypothesis  is  therefore  the  only  admissible  one ; 
that  is,  the  two  triangles  are  mutually  equiangular  and  consequently 
similar. 

34.  Scholium.  Homologous  sides  in  the   two  triangles  are  either 
two  parallel  sides,  or  two  perpendicular  sides ;  and  homologous,  or 
equal,  angles,  are  angles  included  by  homologous  sides. 

PROPOSITION  VIII.— THEOREM. 

35.  If  three  or  more  straight  lines   drawn  through  a  common  point 
intersect  two  parallels,  the  corresponding  segments  of  the  parallels  are 
in  proportion. 

Let    OA,  OB,  OC,  OD,  drawn   through  v 

the  common  point  0,  intersect  the  parallels 
AD  and  ad,  in  the  points  A,  B,  C,  D  and 
a,  b,  c,  d,  respectively ;  then, 

.       .  a/  b 

ab          be         cd 

For,  the  triangle  OAB  is  similar  to  the  tri 
angle  Oab  (25);  OBC  is  similar  to  Obc; 
and  OCD  to  Ocd\  therefore,  we  have 

AB  _OB_BC__gO_ 
ab  "  Ob~~   be         Oc 


/ 


\ 


CD 

cd 


which  includes  the  proportion  that  was  to  be  proved. 


106 


GEOMETRY. 


36.  Scholium.  The  demonstration  is  the  same  whether  the  parallels 
cut  the  system  of  diverging  lines  on  the  same  side,  or  on  opposite 
sides,  of  the  point  0.  Moreover,  the  demonstration  extends  to  any 
corresponding  segments,  as  A  C  and  ac,  BD  and  bd,  etc. ;  and  the 
ratio  of  any  two  corresponding  segments  is  equal  to  the  ratio  of  the 
distances  of  the  parallels  from  the  point  0,  measured  on  any  one  of 
the  diverging  lines. 


PROPOSITION  IX.— THEOREM. 

37.  Conversely,  if  three  or  more  straight  lines  divide  two  parallels 
proportionally,  they  pass  through  a  common  point. 

Let  Aa,  Bb,  Cc,  Dd,  divide  the  parallels 
AD  and  ad  proportionally ;  that  is,  so  that  \  / 

\d        |c      fb  /a_ 

AB_B^_CD 

ab          be         cd 

then,  Aa,  Bb,  etc.,  meet  in  a  common  point. 
For,  let  Aa  and  Cc  meet  in  0 ;  join  Ob. 
Then,  in  order  to  prove  that  Bb   passes 
through  0,  we  have  to  prove  that  Ob  and      / 
Bb  are  in  the  same  straight  line.     Now,  if 

they  are  not  in  the  same  straight  line,  Ob  produced  cuts  AD  in  some 
point  P  differing  from  B ;  and  by  the  preceding  theorem,  we  have 


\ 


AP 
ab 


AC 

ac 


But,  from  the  hypothesis  [1J,  we  have  by  (12), 

AB      AC 


ab 


ac 


whence,  AP  =  AB,  which  is  impossible  unless  P  coincides  with  Bt 
and  Ob  produced  coincides  with  Bb.  Therefore,  Bb  passes  through 
O.  In  the  same  way,  Dd  is  shown  to  pass  through  0. 


BOOK    III. 


107 


c" 


Ff 


PEOPOSITION  X.— THEOREM. 

38.  If  two  polygons  are  composed  of  the  same  number  of  triangles 
similar  each  to  each  and  similarly  placed,  the  polygons  are  similar. 

Let  the  polygon  ABCD,  etc., 
be  composed  of  the  triangles 
ABC,  A  CD,  etc.;  and  let  the 
polygon  A'B'C'D',  etc.,  be  com 
posed  of  the  triangles  AB'C', 
A'C'D',  etc.,  similar  to  ABC, 
ACD,  etc.,  respectively,  and 
similarly  placed  ;  then,  the  polygons  are  similar. 

1st.  The  polygons  are  mutually  equiangular.  For,  the  homolo 
gous  angles  of  the  similar  triangles  are  equal ;  and  any  two  corre 
sponding  angles  of  the  polygons  are  either  homologous  angles  of  two 
similar  triangles,  or  sums  of  homologous  angles  of  two  or  more 
similar  triangles.  Thus  B  =  J5';  BCD  =  BCA  +  ACD  = 
B'C'A  +  A'C'D'  =  B'C'D'-,  etc. 

2d.  Their  homologous  sides  are  proportional.  For,  from  the  simi 
lar  triangles,  we  have 


AB 


BC 


AB'       B'C' 


AC 

AC' 


CD 


AD 


DE 


C'D'       AD'       D'E' 


Therefore,  the  polygons  fulfill  the  two  conditions  of  similarity  (24). 


PROPOSITION  XI.— THEOREM. 

39.  Conversely,  two  similar  polygons  may  be  decomposed  into  the 
same  number  of  triangles  similar  each  to  each  and  similarly  placed. 

Let  ABCD,  etc.,  A'B'C'D', 
etc.,  be  two  similar  polygons. 
From  two  homologous  vertices,  A 
and  A,  let  diagonals  be  drawn  in 
each  polygon  ;  then,  the  polygons 
will  be  decomposed  as  required. 

For,  1st.  We  have,  by  the  definition  of  similar  polygons, 


108 


GEOMETRY. 


Angle  B  =  B',  and  = 


A'B'       B'C' 

therefore,  the  triangles  ABC  and 
A' B'C'  are  similar  (32). 

2d.  Since  ABC  and  A' B'C' 
are  similar,  the  angles  BCA  and 
B'C 'A'  are  equal;  subtracting 
these  equals  from  the  equals  BCD 
and  B'  C'D',  respectively,  there  remain  the  equals  A  CD  and  A'C'D'. 
Also,  from  the  similarity  of  the  triangles  ABC  and  A' B'C',  and 
from  that  of  the  polygons,  we  have 


AC        BC 


AC'       B'C' 


CD 
C'D' 


B'     C' 


therefore,  the  triangles  ACD  and  A'C'D'  are  similar  (32). 

Thus,  successively,  each  triangle  of  one  polygon  may  be  shown  to 
be  similar  to  the  triangle  similarly  situated  in  the  other. 

40.  Scholium.  Two  similar  polygons  may  be  decomposed  into  simi 
lar  triangles,  not  only  by  diagonals,  but  by  lines  drawn  from  any  two 
homologous  points.  Thus,  let  0  be  any  arbitrarily  assumed  point  in 
the  plane  of  the  polygon 
ABCD,  etc.;  and  draw  OA, 
OB,  OC,  etc.  In  the  similar 
polygon  A'B' C'D',  etc.,  draw 
A  0'  making  the  angle 
B'A'O'  equal  to  BAO,  and  F 

B' 0'     making      the     angle 

A'B'  0'  equal  to  AB 0.  The  intersection  0'  of  these  lines,  regarded  as 
a  point  belonging  to  the  polygon  A'B' C'D',  etc.,  is  homologous  to  the 
point  0  of  the  polygon  ABCD,  etc.;  and  the  lines  O'A',  O'B', 
O'C',  etc.,  being  drawn,  the  triangles  O'A'B',  O'B'C',  etc.,  are 
shown  to  be  similar  to  OAB,  OBC,  etc.,  respectively,  by  the  same 
method  as  was  employed  in  the  preceding  demonstration. 

If  the  point  0  is  taken  without  the  polygon,  and  its  homologous 


BOOK    III.  109 

point  0'  found  as  before  by  constructing  the  triangle  0  A'B'  similar 


to  OAB,  the  polygons  will  be  decomposed  into  triangles  partly  addi 
tive  and  partly  subtractive.  Thus  the  polygon  ABODE  is  equal  to 
the  sum  of  the  two  triangles  OBC  and  OCD,  diminished  by  the 
triangles  OBA,  OAE  and  OED;  and  the  polygon  A'E'C'D'E'is 
similarly  decomposed. 

Homologous  lines  in  the  two  polygons  are  lines  joining  pairs  of 
homologous  points,  such  as  OA  and  O'A',  OB  and  O'B',  etc.,  the 
diagonals  joining  homologous  vertices,  etc. ;  and  it  is  readily  shown 
that  any  two  such  homologous  lines  are  in  the  same  ratio  as  any 
two  homologous  sides,  that  is,  in  the  ratio  of  similitude  of  the  poly 
gons  (24). 

41.  Corollary.  Two  similar  polygons  are  equal  when  any  line  in 
one  is  equal  to  its  homologous  line  in  the  other. 

PROPOSITION  XII.— THEOREM. 

42.  The  perimeters  of  two  similar  polygons  are  in  the  same  ratio  as 
any  two  homologous  sides. 

For,  we  have  (see  preceding  figures), 

AB         BC         CD 


A'B'       B'C'       C'D' 

whence  (12), 


AB  +   BC  -f    CD   +  etc.  _    AB          BC 

A'B'  +  B'C'  +  C'D'  +  etc.  ~  A'B'  ~  B'C 


43.  Corollary.  The  perimeters  of  two  similar  polygons  are  in  the 
same  ratio  as  any  two  homologous  lines;  that  is,  in  the  ratio  of 
similitude  of  the  polygons  (40). 

10 


110  GEOMETRY. 

APPLICATIONS. 
PROPOSITION  XIII.— THEOREM. 

44.  If  a  perpendicular  is  drawn  from  the  vertex  of  the  right  angle 
to  the  hypotenuse  of  a  right  triangle : 

1st.    The  two  triangles  thus  formed  are  similar  to  each  other  and  to 
the  whole  triangle ; 

2d.  The  perpendicular  is  a  mean  proportional  between  the  segments 
of  the  hypotenuse ; 

3d.  Each  side  about  the  right  angle  is  a  mean  proportional  between 
the  hypotenuse  and  the  adjacent  segment. 

Let  C  be   the   right   angle  of  the  triangle  /fX. 

ABC,  and  CD  the  perpendicular  to  the  hy-          /     \. 
potenuse ;  then,  A      D  B 

1st.  The  triangles  A  CD  and  CBD  are  simi 
lar  to  each  other  and  to  ABC.     For,  the  triangles  A  CD  and  ABC 
have  the  angle  A  common,  and  the  right  angles,  ADC,  ACB,  equal; 
therefore,  they  are  similar  (26).     For  a  like  reason  CBD  is  similar 
to  ABC,  and  consequently  also  to  A  CD. 

2d.  The  perpendicular  CD  is  a  mean  proportional  between  the 
segments  AD  and  DB.  For,  the  similar  triangles,  A  CD,  CBD,  give 

AD  :  CD  =  CD  :  BD. 

3d.  The  side  A  C  is  a  mean  proportional  between  the  hypotenuse 
AB  and  the  adjacent  segment  AD.  For,  the  similar  triangles,  A  CD, 

ABC,  give 

AB:AC=AC:AD. 

In  the  same  way,  the  triangles  CBD  and  ABC  give, 
AB:BC  =  BC:  BD. 

45.  Corollary  I.  If  all  the  lines  of  the  figure  are  supposed  to  be 
expressed  in  numbers,  being  measured  by  any  cojmmon_  unit,  the 
preceding  proportions  give,  by  (5), 

CD*  =  AD  X  BDt 
(AD, 
C  BD: 


BOOK    III.  Ill 

where  we  employ  the  notation  CD  ,  as  in  algebra,  to  signify  the  pro 
duct  of  CD  multiplied  by  itself,  or  the  second  power  of  CD;  ob 
serving,  however,  that  this  is  but  a  conventional  abbreviation  for 
"second  power  of  the  number  representing  CD"  (8).  It  may  be 
read  "  the  square  of  CD,"  for  a  reason  that  will  appear  hereafter. 

46.  Corollary  II.  By  division,  the  last  two  equations  of  the  pre 
ceding  corollary  give 

AC2^ABXAD      AD 
BC2  ~  AB  X  BD  ~  BD1 

that  is,  the  squares  of  the  sides  including  the  right  angle  are  propor 
tional  to  the  segments  of  the  hypotenuse. 

47.  Corollary  III.  If  from  any  point  C  in  the 
circumference  of  a  circle,  a  perpendicular  CD  is 
drawn  to  a  diameter  AB,  and  also  the  chords  CA, 
CB;  then,  since  A  CB  is  a  right  angle  (II.  59), 

it  follows  that  the  perpendicular  is  a  mean  proportional  between  the 
segments  of  the  diameter ;  and  each  chord  is  a  mean  proportional  be 
tween  the  diameter  and  the  segment  adjacent  to  that  chord. 


PKOPOSITION  XIV.—  THEOEEM. 

48.   The  square  of  the  hypotenuse  of  a  right  triangle  is  equal  to  the 
sum  of  the  squares  of  the  other  two  sides. 
Let  ABC  be  right  angled  at  C;  then, 


BC2. 
For,  by  the  preceding  proposition,  we  have 


AC*  =  ABX  AD,  and  ~BC2  =  AB  X  BD, 
the  sum  of  which  is 

AC*  +  BC2  =  ABX  (AD  +  BD}  =  AB  X  AB  =  AB\ 

49.  Corollary  I.  By  this  theorem,  if  the  numerical  measures  of 
two  sides  of  a  right  triangle  are  given,  that  of  the  third  is  found. 
For  example,  if  AC=  3,  BC=4;  then,  AB  =  ^[S2  +  42]  =  5. 


112  GEOMETRY. 

If  the  hypotenuse,  AB,  and  one  side,  AC,  are  given,  we  have 
J3C'2  =  AB*  —  AC'2;  thus,  if  there  are  given  AB  =  5,  AC  =  3, 
then,  we  find  BC  =  |/[52  —  32]  ==  4. 

50.   Corollary  II.  If  J.  (7  is  the  diagonal  of  a  square 
ABCD,  we  have,  by  the  preceding  theorem, 


whence, 

Z 

and  extracting  the  square  root, 

=  1/2  —  1.41421  +  ad  inf. 


Since  the  square  root  of  2  is  an  incommensurable  number,  it  follows 
that  the  diagonal  of  a  square  is  incommensurable  with  its  side. 

51.  Definition.  The  projection  of  a  point  A  A 

upon  an  indefinite  straight  line  XY  is  the  foot 
P  of  the  perpendicular  let  fall  from  the  point 
upon  the  line.  j> 

The  projection  of  a  finite  straight  line  AB 

upon  the  line  XY  is  the  distance  PQ  between  the  projections  of  the 
extremities  of  AB. 

If  one  extremity  B  of  the  line  AB  is  in  the 
line  XY,  the  distance  from  B  to  P  (the  projec 
tion  of  A)  is  the  projection  of  AB  on  XY;  for 
the  point  B  is  in  this  case  its  own  projection. 

PROPOSITION  XV.— THEOREM. 

52.  In  any  triangle,  the  square  of  the  side  opposite  to  an  acute  angle 
is  equal  to  the  sum  of  the  squares  of  the  other  two  sides  diminished  by 
twice  the  product  of  one  of  these  sides  and  the  projection  of  the  other 
upon  that  side.  (?  pi  i 

Let  C  be  an  acute  angle  of  the  triangle  ABC, 
Pthe  projection  of  A  upon  BC  by  the  perpen 
dicular  AP,  PC  the  projection  of  A  C  upon  BC; 
then, 


BOOK    III.  113 


"  =  BCZ  +  AC2  -  2BC  X  PC. 

For,  if  P  falls  on  the  base,  as  in  Fig.  1,  we 
,  j* 

have 

PB  =  BC  —  PC, 

and  if  P  falls  upon  the  base  produced,  as  in  Fig.  2,  we  have 


but  in  either  case  the  square  of  PB  is,  by  a  theorem  of  algebra,  * 
PB2  =  WC2  +  PC*  —  2BC  X  PC. 

Adding  AP*  to  both  members  of  this  equality,  and  observing  that 
by  the  preceding  theorem,  PZ?2  -f  ZF  =  AB\  and  PC*  -j-  ZP2  = 
'2,  we  obtain 


PROPOSITION  XVI.— THEOEEM. 

53.  In  an  obtuse  angled  triangle,  the  square  of  the  side  opposite  to 
the  obtuse  angle  is  equal  to  the  sum  of  the  squares  of  the  other  two  sides, 
increased  by  twice  the  product  of  one  of  these  sides  and  the  projection 
of  the  other  upon  that  side. 

Let  Cbe  the  obtuse  angle  of  the  triangle  ABC, 
P  the  projection  of  A  upon  BC  (produced) ;  then, 


'   PC.  PC  B 

For,  since  P  can  only  fall  upon  BC  produced,  ACB  being  an 
obtuse  angle,  we  shall  in  all  cases  have 

PB  =  BC-i-  PC, 

and  the  square  of  PB  will  be,  by  an  algebraic  theorem,  f 
PB2  =  BC2  +  P<72  +  2BC  X  PC. 


Adding  AP   to  both  members,  we  obtain 

AB2  =  &C2  -f  AC2  -f  2BC  X  PC. 

*  (x  —  y}z  or  (y  —  x}2  =  x*  -|-  2/2  —  2xy. 
f  (x  -f  y)2  =  x2  +  y2  +  2zy. 

10.  H 


114 


GEOMETRY. 


54.  Corollary.  From  the  preceding  three  theorems,  it  follows  that 
an  angle  of  a  triangle  is  acute,  right  or  obtuse,  according  as  the 
square  of  the  side  opposite  to  it  is  less  than,  equal  to,  or  greater  than, 
the  sum  of  the  squares  of  the  other  two  sides. 


PROPOSITION  XVII.— THEOEEM. 

55.  If  through  a  fixed  point  within  a  circle  any  chord  is  drawn,  the 
product  of  its  two  segments  has  the  same  value,  in  whatever  direction  the 
chord  is  drawn. 

Let  P  be  any  fixed  point  within  the  circle  0, 
AB  and  A'B'  any  two  chords  drawn  through  P; 
then, 

PA  X  PB  =  PA'  X  PB'. 

For,  join  AB'  and  A'B.  The  triangles  APB't 
A'PB,  are  similar,  having  the  angles  at  P  equal, 
and  also  the  angles  A  and  A'  equal  (II.  58) ;  therefore, 


whence  (5), 


PA  :  PA  =  PB'  :  PB, 


PAXPB  =  PA'  X  PB'. 


56.  Corollary.  If  AB  is  the  least  chord,  drawn 
through  P  (II.  20),  then,  since  it  is  perpendicular 
to  OP,  we  have  PA  =  PB  (II.  15),  and  hence 
PA2  =  PA'  X  PB' ;  that  is,  either  segment  of  the 
least  chord  drawn  through  a  fixed  point  is  a  mean 

proportional  between  the  segments  of  any  other  chord  drawn  through 
that  point. 

57.  Scholium.  If  a  chord  constantly  passing  through  a  fixed  point 
P,  be  conceived  to  revolve  upon  this  point  as  upon  a  pivot,  one  seg 
ment  of  the  chord  increases  while  the  other  decreases,  but  their 
product  being  constant  (being  always  equal  to  the  square  of  half  the 
least  chord),  the  two  segments  are  said  to  vary  reciprocally,  or  to  be 
reciprocally  proportional  (2). 


BOOK     III.  115 


PROPOSITION  XVIII— THEOREM. 

58.  If  through  a  fixed  point  without  a  circle  a  secant  is  drawn,  the 
product  of  the  whole  secant  and  its  external  segment  has  the  same  value, 
in  whatever  direction  the  secant  is  drawn. 

Let  P  be  any  fixed  point  without  the  circle  0, 
PAB  and  PA'B'  any  two  secants  drawn  through  P; 
then, 

PA  X  PB  =  PA'  X  PB'. 

For,  join  AB'  and  A'B.  The  triangles  APBf, 
A'PB,  are  similar,  having  the  angle  at  P  common, 
and  also  the  angles  B  and  B'  equal  (II.  58)  ;  there 
fore, 

PA:PA'  =  PB'iPB, 
whence  (5), 

PA  X  PB  =  PA'  X  PB'. 

59.  Corollary.  If  the  line  PAB,  constantly  passing  through  the 
fixed  point  P,  be  conceived  to  revolve  upon  P,  as  upon  a  pivot,  and 
to  approach  the  tangent  PT,  the  two  points  of  intersection,  A  and  B, 
will  approach  each  other ;  and  when  the  line  has  come  into  coinci 
dence  with  the  tangent,  the  two  points  of  intersection  will  coincide 
in  the  point  of  tangency  T.     The  whole  secant  and  its  external  seg 
ment  will  then  both  become  equal  to  the  tangent  PT;   therefore, 
regarding  the  tangent  as  a  secant  whose  two  points  of  intersection 
are  coincident  (II.  28),  we  shall  have 

PT2  =  PA'  X  PS'; 

that  is,  if  through  a  fixed  point  without  a  circle  a  tangent  to  the  circle 
is  drawn,  and  also  any  secant,  the  tangent  is  a  mean  proportional  be 
tween  the  whole  secant  and  its  external  segment. 

60.  Scholium  I.  When  a  secant,  constantly  passing  through  a  fixed 
point,  changes  its  direction,  the  whole  secant  and  its  external  seg 
ment  vary  reciprocally,  or  they  are  reciprocally  proportional,  since 
their  product  is  constant  (2). 

61.  Scholium  II.  The  analogy  between  the  two  preceding  proposi 
tions  is  especially  to  be  remarked.     They  may,  indeed,  be  reduced 
to  a  single  proposition  in  the  following  form :  If  through  any  fixed 


116 


GEOMETRY. 


point  in  the  plane  of  a  circle  a  straight  line  is  drawn  intersecting  the 
circumference,  the  product  of  the  distances  of  the  fixed  point  from  the 
two  points  of  intersection  is  constant. 


PROPOSITION  XIX.—  THEOREM. 

62.  In  any  triangle,  if  a  medial  line  is  drawn  from  the  vertex  to  the 
base  : 

1st.  The  sum  of  the  squares  of  the  two  sides  is  equal  to  twice  the 
square  of  half  the  base  increased  by  twice  the  square  of  the  medial  line; 

2d.  The  difference  of  the  squares  of  the  two  sides  is  equal  to  twice 
the  product  of  the  base  by  the  projection  of  the  medial  line  on  the  base. 

In  the  triangle  ABC,  let  D  be  the  middle 
point  of  the  base  B  C,  AD  the  medial  line  from 
A  to  the  base,  P  the  projection  of  A  upon  the 
base,  DP  the  projection  of  AD  upon  the  base  ; 
then, 

1st.  AB2  +  AC*  =  2BD2  +  2AD*  ; 

2d.   IS9  —  AC'2  =  2BC  X    DP. 

For,  if  AB  >  AC,  the  angle  ADB  will  be  obtuse  and  ADO  will 
be  acute,  and  in  the  triangles  ABD,  ADC,  we  shall  have,  by  (53) 
and  (52). 

2  =  BD*  +  AD'  +  2BD  X  DP, 


AC2=I)C2  +  AD2  —  2DC  X  DP.< 
Adding  these  equations,  and  observing  that  BD  —  DC,  we  have 

1st.      AB*  +  AC2  =  2BD*  +  2AD2. 
Subtracting  the  second  equation  from  the  first,  we  have 
'AC2  =  2(BD  +  DC)  X  DP-, 
2d.      AB'  —  AC'1  =  2BC  X  DP. 


that  is, 


BOOK    III. 


117 


63.   Corollary  I.  In  any  quadrilateral,  the  sum  of  the  squares  of 
the  four  sides  is  equal  to  the  sum  of  the  squares 
of  the  diagonals  plus  four  times  the  square  of 
the  line  joining  the  middle  points  of  the  diag- 
oiials. 

For,  let  E  and  F  be  the  middle  points  of  the 
diagonals  of  the  quadrilateral  ABCD',  join 
EF,  EB,  ED.  Then,  by  the  preceding  theorem, 
we  have  in  the  triangle  ABC, 


AB  +  BC  =  2AE 
and  in  the  triangle  ADC, 

CD2  +  DA2  =  2AE2  +  2DE\ 
whence,  by  addition, 

AB*  +  BC*  _j_  CD2  +  DA2  =  4AE2  +  2  (BE 
Now,  in  the  triangle  BED,  we  have 

BE2  +  DE2  =  2BF2  +  2EF2-, 
therefore, 

IB2  +  BC2  +  CD2  +  DA2  =  4AE2  +  4BF2  +  4EF2. 

But  4AE2  =  (2AE?  =  AC2,  and  4BF2  =  (2BF?  =  BD2; 
hence,  finally, 

AB2  +  BC2  +  CD2  +  DA2  =  AC*  +  BD2  +  4£F2. 

64.  Corollary  II.  In  a  parallelogram,  the  sum  of  the  squares  of 
the  four  sides  is  equal  to  the  sum  of  the  squares  of  the  diagonals. 
For  if  the  quadrilateral  in  the  preceding  corollary  is  a  parallelo 
gram,  the  diagonals  bisect  each  other,  and  the  distance  EF  is  zero. 


PKOPOSITION  XX.— THEOEEM. 

65.  In  any  triangle,  the  product  of  two  sides  is  equal  to  the  product 
of  the  diameter  of  the  circumscribed  circle  by  the  perpendicular  let  fall 
upon  ike  third  side  from  the  vertex  of  the  opposite  angle. 


118 


GEOMETRY. 


Let  AB,  A  C,  be  two  sides  of  a  triangle  AB  C, 
AD  the  perpendicular  upon  BC,  AE  the  di 
ameter  of  the  circumscribed  circle ;  then, 

AB  X  AC=AEX  AD. 


7 C 


For,  joining  CE,  the  angle  ACE  is  a  right 
angle  (II.  59),  and  the  angles  E  and  B  are  equal  (II.  58) ;  there 
fore,  the  right  triangles  AEC,  ABD,  are  similar,  and  give 

AB  :  AE  =  AD  :  AC, 
whence,  AB  X  A  C  =  AE  X  AD. 


PKOPOSITION  XXI.— THEOEEM. 

66.  In  any  triangle,  the  product  of  two  sides  is  equal  to  the  product 
of  the  segments  of  the  third  side  formed  by  the  bisector  of  the  opposite 
angle  plus  the  square  of  the  bisector. 

Let  AD  bisect  the  angle  A  of  the 
triangle  ABC;  then, 


,AB  X  AC=DB  X 


DA\ 


For,  circumscribe  a  circle  about 
ABC,  produce  AD  to  meet  the  cir 
cumference  in  E,  and  join  CE.  The 
triangles  ABD,  AEC,  are  similar,  and  give 

AB:AE=DA:AC, 


whence    AB  X  A  C  =  AE  X  DA  =  (DE  +  DA}  X  DA 
=  DE  X  DA  +  DA\ 

Now,  by  (55),  we  have  DE  X  DA  =  DB  X  DC,  and  hence 
ABX  AC=DBX  DC+DA\ 

67.   Corollary.  If  the  exterior  angle  BAF  is  bisected  by  AD',  the 
same  theorem  holds,  except  that  plus  is  to  be  changed  to  minus. 


BOOK    III. 


119 


For,  producing  D'A  to  meet  the  circumference  in  E',  and  joining 
CE',  the  triangles  ABD',  AE'C,  are  similar,  and  give 

AS:AE'  =  AD'  -.AC, 

whence  AS  X  AC  =  AE'  X  AD'  =  (D'Er  —  D'A}  X  D'A 
=  D'E'  X  D'A  —  WA\ 

or,  by  (58),  AB  X  AC=  D'B  X  D'C— 


PKOBLEMS    OF    CONSTBUCTION. 

PKOPOSITION  XXII.— PKOBLEM. 

68.   To  divide  a  given  straight  line  into  parts  proportional  to  given 
straight  lines. 

Let  it  be  required  to  divide  AB  into  parts 
proportional  to  M,  N  and  P.  From  A  draw 
an  indefinite  straight  line  AX,  upon  which  lay 
off  AC=  M,  CD  =  N,  DE  =  P,  join  EB, 
and  draw  C.F,  DG,  parallel  i^EB',  then  AFt 
FG,  GB,  are  proportional  to  M,  N,  P  (18). 


69.  Corollary.  To  divide  a  given  straight  line  AB  into  any  num 
ber  of  equal  parts,  draw  an  indefinite  line  AX,  upon 
which  lay  off  the  same  number  of  equal  distances, 
each  distance  being  of  any  convenient  length ;  through 
M  the  last  point  of  division  on  AX  draw  MB,  and 
through  the  other  points  of  division' of  AX  draw  par 
allels  to  MB,  which  will  divide  AB  into  the  required 
number  of  equal  parts.  This  follows  both  from  the 
theory  of  proportional  lines  and  from  (I.  125). 


120 


GEOMETRY. 


MNP 


PKOPOSITION  XXIII.— PROBLEM. 

70.  To  find  a  fourth  proportional  to  three  given  straight  lines. 
Let  it  be  required  to  find  a  fourth  propor 
tional  to  M,  N  and  P.     Draw  the  indefinite 

lines  AX,  AY,  making  any  angle  with  each 
other.  Upon  AX  lay  off  AB  =  M,AD  =  N-, 
and  upon  AY  lay  off  AC  =  P;  join  BC,  and 
draw  DE  parallel  to  BC',  then  AE  is  the  re 
quired  fourth  proportional. 
For,  we  have  (15), 

AB  :  AD  =  AC:  AE,  or  M :  N=P:AE. 

71.  Corollary.  If  AB  =  M,  and  both  AD  and  J.Care  made  equal 
to  N,  AE  will  be  a  third  proportional  to  If  and  N;  for  we  shall  have 

M:N=N:AE. 


PKOPOSITION  XXIV.— PROBLEM. 

72.   To  find  a  mean  proportional  between  two  given  straight  lines. 

Let  it  be  required  to  find  a  mean  proportional 
between  M  and  N.  Upon  an  indefinite  line  lay 
off  AB  =  M,  BC  =  N;  upon  AC  describe  a 
semi-circumference,  and  at  B  erect  a  perpen 
dicular,  BD,  to  AC.  Then  BD  is  the  required 
mean  proportional  (47). 

Second  method.  Take  AB  equal  to  the  greater 
line  M,  and  upon  it  lay  off  BC  =  N.  Upon 
AB  describe  a  semi-circumference,  erect  CD  per 
pendicular  to  AB  and  join  BD.  Then  BD  is 
the  required  mean  proportional  (47). 


73.  Definition.  When  a  given  straight  line  is  divided  into  two 
segments  such  that  one  of  the  segments  is  a  mean  proportional 
between  the  given  line  and  the  other  segment,  it  is  said  to  be  divided 
in  extreme  and  mean  ratio. 

Thus  AB  is  divided  in  extreme  and       I- J + — j 

v  /  -A  C        Jf 

mean   ratio   at   C,   if   AB  :   AC  = 
AC:  CB. 


BOOK    III.  121 

If  C'  is  taken  in  BA  produced  so  that  AB  :  AC'  =  AC'  :  C'B, 
then  AB  is  divided  at  C",  externally,  in  extreme  and  mean  ratio. 


PKOPOSITION  XXV.— PEOBLEM. 

74.   To  divide  a  given  straight  line  in  extreme  and  mean  ratio. 

Let  AB  be  the  given  straight  line.     At  B  erect  the  perpendicular 
BO  equal  to  one  half  of  AB. 
With  the  centre  0  and  radius 
OB,  describe   a   circumference, 
and  through  A  and  0  draw  A  0 

cutting  the  circumference  first      \- 

in  D  and  a  second  time  in  D'. 
Upon  AB  lay  off  AC  =  AD,  and  upon  BA  produced  lay  off 
AC'  =  AD'.  Then  AB  is  divided  at  C  internally,  and  at  C'  exter 
nally,  in  extreme  and  mean  ratio. 

For,  1st,  we  have  (59), 

AD'  :AB  =  AB:AD  or  AC,  [1] 

whence,  by  division  (10), 

AD'  —  AB:AB  =  AB  —  AC:  AC, 
or,  since  DD'  =  2 OB  =  AB,  and  therefore  AD'  —  AB  =  AD'  — 

=  CB:AC, 


and,  by  inversion  (7), 

AB:AC=AC:CB', 

that  is,  AB  is  divided  at  C,  internally,  in  extreme  and  mean  ratio. 
2d.  The  proportion  [1]  gives  by  composition  (10), 

AD'  +  AB  :  AD'  =  AB  +  AD  :  AB, 

or,  since  AD'  =  AC',  AD'  +  AB  =  C'B,  AB  +  AD  =  DD'  + 
AD  =  AD'  =  AC', 


and,  by  inversion, 

AB:AC'  =  AC'i  C'B-, 


that  is,  AB  is  divided  at  C',  externally,  in  extreme  and  mean  ratio. 
11 


122  GEOMETRY. 

75.  Scholium.  Since  OD  =  OD'  =  —,  we  have 

2 

AC  =  AO-  —         AC'  =  AO  +  ~ 

2  '  2  ' 

But  the  right  triangle  A  OB  gives 

hllT/  =  =  ^-4' 
whence,  extracting  the  square  root, 


2 
Therefore, 


76.  Definitions.  When  a  straight  line  is  divided  internally  and 
externally  in  the  same  ratio,  it  is  said  to  be  divided  harmonically. 

Thus,  AB  is  divided  harmonically 
at  C  and  D,ii  CA  :  CB  =  DA  :  DB-,       I  -  1  -  1  -  I 

-4  G        B  D 

that  is,  if  the  ratio  of  the  distances 

of  C  from  A  and  B  is  equal  to  the  ratio  of  the  distances  of  D  from 

A  and  B. 

Since  this  proportion  may  also  be  written  in  the  form 


the  ratio  of  the  distances  of  A  from  C  and  D  is  equal  to  the  ratio 
of  the  distances  of  B  from  C  and  D ;  consequently  the  line  CD  is 
divided  harmonically  at  A  and  B. 

The  four  points  A,  B,  C,  D,  thus  related,  are  called  harmonic 
points,  and  A  and  jB  are  called  conjugate  points,  as  also  C  and  D. 

PROPOSITION  XXVI.— PROBLEM. 

77.   To  divide  a  given  straight  line  harmonically  in  a  given  ratio. 

Let  it  be  required  to  divide  AB 
harmonically  in  the  ratio  of  M  to  N.  m , 

Upon  the  indefinite  line  AX,  lay 
off  AE  =  M,  and  from  E  lay  off  EF 
and  EG,  each  equal  to  N;  join  FB, 
GB-,  and  draw  EC  parallel  to  FB, 
ED  parallel  to  GB. 


BOOK     III.  123 

Then,  by  the  construction  we  have  (17), 
M  CA  DA 
N~  CB~  DB1 

therefore,  by  the  definition  (76),  AB  is  divided  harmonically  at  C 
and  D,  and  in  the  given  ratio. 

78.  Scholium.  If  the  extreme  points  A  and  D  are  given,  and  it  is 
required  to  insert  their  conjugate  harmonic  points  B  and  C,  the  har 
monic  ratio  being  given  =  M :  N,  we  take  on  AX,  as  before,  AE  = 
M  and  EF  =  EG  =  N,  join  ED,  and  draw  GB  parallel  to  ED, 
which  determines  B ;  then,  join  FB  and  draw  EC  parallel  to  FBy 
which  determines  (7. 

Also  if,  of  four  harmonic  points  A,  B,  C,  D,  any  three  are  given, 
the  fourth  can  be  found. 


PKOPOSITION  XXVII.— PKOBLEM. 

79.  To  find  the  locus  of  all  the  points  whose  distances  from  two  given 
points  are  in  a  given  ratio. 

Let  A  and  B  be  the  given  points,  and  let  the  given  ratio  be  M  :  N. 
Suppose  the  problem  solved,  and 
that  P  is  a  point  of  the  required 
locus.  Divide  AB  internally  at 
C  and  externally  at  D,  in  the  ratio 
M :  N,  and  join  PA,  PB,  PC,  PD. 
By  the  condition  imposed  upon  P 
we  must  have 

PA:PB  =  M:  N=  CA  :  CB  =  D A  :  DB; 

therefore,  PC  bisects  the  angle  APB,  and  PD  bisects  the  exterior 
angle  BPE  (23).  But  the  bisectors  PC  and  PD  are  perpendicular  to 
each  other  (I.  25) ;  therefore,  the  point  P  is  the  vertex  of  a  right 
angle  whose  sides  pass  through  the  fixed  points  C  and  D,  and  the 
locus  of  P  is  the  circumference  of  a  circle  described  upon  CD  as 
a  diameter  (II.  59,  97).  Hence,  we  derive  the  following 

Construction.  Divide  AB  harmonically,  at  C  and  D,  in  the  given 
ratio  (77),  and  upon  CD  as  a  diameter  describe  a  circumference. 
This  circumference  is  the  required  locus. 


124 


GEOMETRY. 


PKOPOSITION  XXVIII.— PKOBLEM. 

80.   On  a  given  straight  line,  to  construct  a  polygon  similar  to  a  given 
polygon. 

Let  it  be  required  to  construct 
upon  A'B'  a  polygon  similar  to 
ABCDEF. 

Divide  ABCDEF  into  tri 
angles  by  diagonals  drawn  from 
A.  Make  the  angles  B'A'C' 

and  A'B'C'  equal  to  BACtmd  AB C  respectively ;  then,  the  triangle 
A'B' C'  will  be  similar  to  ABC  (25).  In  the  same  manner  construct 
the  triangle  A'D'C'  similar  to  ADC,  A'E'D'  similar  to  AED,  and 
A'E'F'  similar  to  AEF.  Then,  A'B'C'D'E'F'  is  the  required 
polygon  (38). 


PKOPOSITION  XXIX.— PKOBLEM. 

81.   To  construct  a  polygon  similar  to  a  given  polygon,  the  ratio  of 
similitude  of  the  two  polygons  being  given. 

Let  ABODE  be  the  given 
polygon,  and  let  the  given  ratio 
of  similitude  be  M :  N. 

Take  any  point  0,  either 
within  or  without  the  given 
polygon,  and  draw  straight  lines 
from  0  through  each  of  the 
vertices  of  the  polygon.  Upon 

any  one  of  these  lines,  as  OA,  take  OA'  a  fourth  proportional  to 
M,  N,  and  OA,  that  is,  so  that 

M:N=  OA:  OA'. 

In  the  angle  .401?  draw  A'B'  parallel  to  AB-,  then,  in  the  angle 
BOG,  B'Cf  parallel  to  BC,  and  so  on.  The  polygon  A'B'C'D'E' 
will  be  similar  to  ABCDE;  for  the  two  polygons  will  be  composed 


BOOK    III. 


125 


of  the  same  number  of  triangles,  additive  or  subtractive,  similarly 
placed;  and  their  ratio  of  similitude  will  evidently  be  the  given 
ratio  M  :N.  (40). 

82.  Scholium.  The  point  0  in  the  preceding  construction  is  called 
the  centre  of  similitude  of  the  two  polygons. 
11* 


BOOK   IV. 

COMPARISON  AND  MEASUREMENT  OF  THE  SURFACES  OF 
RECTILINEAR  FIGURES. 

1.  DEFINITION.  The  area  of  a  surface  is  its  numerical  measure, 
referred  to  some  other  surface  as  the  unit ;  in  other  words,  it  is  the 
ratio  of  the  surface  to  the  unit  of  surface  (II.  43). 

The  unit  of  surface  is  called  the  superficial  unit.     The  most  con 
venient  superficial  unit  is  the  square  whose  side  is  the  linear  unit. 

2.  Definition.  Equivalent  figures  are  those  whose  areas  are  equal. 


PKOPOSITION  I.— THEOKEM. 

3.   Two  rectangles  having  equal  altitudes  are  to  each  other  as  their 
bases. 

Let  ABCD,  AEFD,  be  two  rectangles  hav-      D  F 

ing  equal  altitudes,  AB  and  AE  their  bases ; 
then,  A 


ABCD 
AEFD 


AB 
AE 


Suppose  the  bases  to  have  a  common  meas 
ure  which  is  contained,  for  example,  7  times  in 
AB,  and  4  times  in  AE\  so  that  if  AB  is 
divided  into  7  equal  parts,  AE  will  contain  4  of  these  parts  ;  then, 
we  have 


AE~  4 

If,  now,  at  the  several  points  of  division  of  the  bases,  we  erect 
perpendiculars  to  them,  the  rectangle  ABCD  will  be  divided  into  7 

126 


BOOK    IV 


127 


equal  rectangles  (I.  120),  of  which  AEFD  will  contain  4  ;  conse 
quently,  we  have 

ABCD       7 

AEFD  ~  4 
and  therefore 


ABCD  = 
AEFD  ~  AE 

The  demonstration  is  extended  to  the  case  in  which  the  bases  are 
incommensurable,  by  the  process  already  exemplified  in  (II.  51) 
and  (III.  15). 

4.  Corollary.  Since  AD  may  be  called  the  base,  and  AB  and  AE 
the  altitudes,  it  follows  that  two  rectangles  having  equal  bases  are  to 
each  other  as  their  altitudes. 

Note.  In  these  propositions,  by  "  rectangle"  is  to  be  understood 
"  surface  of  the  rectangle." 


PEOPOSITION  II.— THEOEEM. 

5.  Any  two  rectangles  are  to  each  other  as  the  products  of  their  bases 
by  their  altitudes. 

Let  E  and  R'  be  two  rectangles, 
k  and  k'  their  bases,  h  and  hr  their 
altitudes;  then, 


?L 

R' 


kXh 
k'  X  h'' 


For,  let  S  be  a  third  rectangle 

having  the  same  base  k  as  the  rec 

tangle  R,  and  the  same  altitude  h'  as  the  rectangle  R'  ;  then  we 

have,  by  (4)  and  (3), 


8  ~  hf        R'  ~  k' 
and  multiplying  these  ratios,  we  find  (III.  14), 

72  _      k  X  h 
R'~  k'  X  h'' 

6.  Scholium.  It  must  be  remembered  that  by  the  product  of  two 


128 


GEOMETRY. 


lines,  is  to  be  understood  the  product  of  the  numbers  which  represent 
them  when  they  are  measured  by  the  linear  unit  (III.  8). 


PROPOSITION  III.— THEOREM. 

7.  The  area  of  a  rectangle  is  equal  to  the  product  of  its  base  and 
altitude. 

Let  H  be  any  rectangle,  k  its  base  and 
h  its  altitude  numerically  expressed  in 
terms  of  the  linear  unit;  and  let  Q  be 
the  square  whose  side  is  the  linear  unit ; 
then,  by  the  preceding  theorem, 

R_kyji_k 

Q~~l  X  l~ 

n 

But  since  Q  is  the  unit  of  surface,  —  =  the  numerical  measure,  or 

<b 

area,  of  the  rectangle  R  (1) ;  therefore, 

Area  of  R  =  k  X  h. 

8.  Scholium  I.  When  the  base  and  altitude  are  exactly  divisible 
by  the  linear  unit,  this  proposition   is   rendered 

evident  by  dividing  the  rectangle  into  squares  each 
equal  to  the  superficial  unit.  Thus,  if  the  base 
contains  7  linear  units  and  the  altitude  5,  the  rec 
tangle  can  obviously  be  divided  into  35  squares 
each  equal  to  the  superficial  unit;  that  is,  its  area1'—  5X7.  The 
proposition,  as  above  demonstrated,  is,  however,  more  general,  and 
includes  also  the  cases  in  which  either  the  base,  or  the  altitude,  or 
both,  are  incommensurable  with  the  unit  of  length. 

9.  Scholium  II.  The  area  of  a  square  being  the  product  of  two 
equal  sides,  is  the  second  power  of  a  side.     Hence  it  is,  that  in  arith 
metic  and  algebra,  the  expression  "square  of  a  number"  has  been 
adopted  to  signify  "second  power  of  a  number." 

We  may  also  here  observe  that  many  writers  employ  the  expres 
sion  "  rectangle  of  two  lines"  in  the  sense  of  "  product  of  two  lines," 
because  the  rectangle  constructed  upon  two  lines  is  measured  by  the 
product  of  the  numerical  measures  of  the  lines. 


BOOK    IV. 


129 


PKOPOSITION  IV.— THEOKEM. 

10.   The  area  of  a  parallelogram  is  equal  to  the  product  of  its  base 
and  altitude. 

Let  ABCD  be  a  parallelogram,  Jc  the     *    D          E     c 
numerical  measure  of  its  base  AB,  h 
that  of  its  altitude  AF;  and  denote  its 
area  by  S;  then, 


For,  let  the  rectangle  ABEF  be  con 
structed  having  the  same  base  and  alti 
tude  as  the  parallelogram ;  the  upper  bases  of  the  two  figures  will  be 
in  the  same  straight  line  FC  (I.  58).  The  right  triangles  AFD  and 
BEG  are  equal,  having  AF  =  BE,  and  AD  =  BC  (I.  83).  If 
from  the  whole  figure  ABCFwe  take  away  the  triangle  AFD,  there 
remains  the  parallelogram  ABCD;  and  if  from  the  whole  figure  we 
take  away  the  triangle  BEC,  there  remains  the  rectangle  ABEF; 
therefore  the  surface  of  the  parallelogram  is  equal  to  that  of  the 
rectangle.  But  the  area  of  the  rectangle  is  Ic  X  h  (7) ;  therefore 
that  of  the  parallelogram  is  also  k  X  h ;  that  is  S  =  k  X  h. 

11.  Corollary  I.  Parallelograms  having  equal  bases  and  equal  alti 
tudes  are  equivalent. 

12.  Corollary  II.    Parallelograms  having  equal  altitudes  are  to 
each  other  as  their  bases ;  parallelograms  having  equal  bases  are  to 
each  other  as  their  altitudes;  and  any  two  parallelograms  are  to 
each  other  as  the  products  of  their  bases  by  their  altitudes. 


PEOPOSITION  V.— THEOREM. 

13.   The  area  of  a  triangle  is  equal  to  half  the  product  of  its  base 
and  altitude. 

Let  AB  (7 be  a  triangle,  k  the  numerical  meas- 

ure  of  its  base  BC,  h  that  of  its  altitude  AD;          /  ^^^      /  -,h 
and  S  its  area ;  then, 

8  =  J  fc  X  h. 


For,  through  A  draw  AE  parallel  to  CB,  and  through  B  draw  BE 

I 


130  GEOMETRY. 

parallel  to  CA.  The  triangle  ABC  is  one-half  the  parallelogram 
AEBC  (1. 105)  ;  but  the  area  of  the  parallelogram  ==  k  X  h ;  there 
fore,  for  the  triangle,  we  have  tS  =  \  k  X  h. 

14.  Corollary  I.  A  triangle  is  equivalent  to  one-half  of  any  par 
allelogram  having  the  same  base  and  the  same  altitude. 

15.  Corollary  II.  Triangles  having  equal  bases  and  equal  altitudes 
are  equivalent. 

16.  Corollary  III.   Triangles  having  equal  altitudes  are  to  each 
other  as  their  bases ;  triangles  having  equal  bases  are  to  each  other 
as  their  altitudes ;  and  any  two  triangles  are  to  each  other  as  the 
products  of  their  bases  by  their  altitudes. 


PKOPOSITION  VI.—  THEOREM. 

17.   The  area  of  a  trapezoid  is  equal  to  the  product  of  its  altitude  by 
half  the  sum  of  its  parallel  bases. 

Let  ABCD  be  a  trapezoid;  MN  =  h,  its  al-  4    x          D 

titude;  AD  =  a,  BC  =  b,  its  parallel  bases;       E 

and  let  S  denote  its  area  ;  then,  / 

/  ' 

B  N 


For,  draw  the  diagonal  A  C.  The  altitude  of  each  of  the  triangles 
AD  Gaud  ABC  is  equal  to  h,  and  their  bases  are  respectively  a  and 
b  ;  the  area  of  the  first  is  I  a  X  h,  that  of  the  second  is  £  b  X  h  ;  and 
the  trapezoid  being  the  sum  of  the  two  triangles,  we  have 

S  =  $a  X  h  +  $b  X  h  =  }(a  +  i)  X  h. 

18.  Corollary.  The  straight  line  EF,  joining  the  middle  points  of 
AB  and  DC,  being  equal  to  half  the  sum  of  AD  and  BC  (I.  124), 
the  area  of  the  trapezoid  is  equal  to  the  product  MN  X  EF. 

19.  Scholium.  The  area  of  any  polygon  may  be  found  by  finding 
the  areas  of  the  several  triangles  into  which  it  may  be  decomposed 
by  drawing  diagonals  from  any  vertex. 

The  following  method,  however,  is  usually  preferred,  especially  in 
surveying.  Draw  the  longest  diagonal  AD  of  the  proposed  polygon 


BOOK     IV. 


131 


ABCDEF]  and  upon  AD  let  fall  the  per 
pendiculars  BM,  CN,  EP,  FQ.  The  poly 
gon  is  thus  decomposed  into  right  triangles 
and  right  trapezoids,  and  by  measuring  the 
lengths  of  the  perpendiculars  and  also  of  the 
distances  AM,  MN,  ND,  AQ,  QP,  PD,  the 
bases  and  altitudes  of  these  triangles  and 

trapezoids  are  known.  Hence  their  areas  can  be  computed  by  the 
preceding  theorems,  and  the  sum  of  these  areas  will  be  the  area  of 
the  polygon. 


PKOPOSITION  VII.— THEOKEM. 

20.  Similar  triangles  are  to  each  other  as  the  squares  of  their  homolo 
gous  sides. 

Let  ABC,  A'B'C'  be  similar  tri 
angles;  then, 

ABC         BC* 


777/2 


A'B'C'       B'C1 

Let    AD,  A'D',   be   the   altitudes. 
By  (16),  we  have 


D          C 


ABC          BCXAD          BC         AD 
—  -~.  ~.  X 


A'B'C'       B'C'  X  A'D'       B'C'       A'D' 

But  the  homologous  lines  AD,  A'D',  are  in  the  ratio  of  similitude 
of  the  triangles  (III.  29) ;  that  is, 


therefore, 


AD  _     BC 
A'D' ~^  B'C' 


ABC  _  BC        BC  _  BC2 
A'B'C'  ~~  B'C'       B'C'    -  BrC''i 


21.   Corollary.  If  we  had  put  the  ratio  AD  :  A'D'  in  the  place  of 
the  ratio  BC:  B'C',  we  should  have  found 

ABC         AD2 


A'B'C'    ~  A'D'*' 


132  GEOMETRY. 

and  in  general,  we  may  conclude  that  the  surfaces  of  two  similar  tri 
angles  are  as  the  squares  of  any  two  homologous  lines;  or,  again,  the 
ratio  of  the  surfaces  of  two  similar  triangles  is  the  square  of  the  ratio 
of  similitude  of  the  triangles. 


PKOPOSITION  VIII.— THEOREM. 

22.   Two  triangles  having  an  angle  of  the  one  equal  to  an  angle  of 
the  other  are  to  each  other  as  the  products  of  the  sides  including  the 


Two  triangles  which  have  an  angle  of  the  one 
equal  to  an  angle  of  the  other  may  be  placed  with 
their  equal  angles  in  coincidence.  Let  ABC,  ADE, 
be  the  two  triangles  having  the  common  angle  A ; 
then, 

ABC_ABX  AC 
ADE  ~  AD  X  AE 

For,  join  BE.  The  triangles  ABC,  ABE,  having  the  common 
vertex  B,  and  their  bases  A  C,  AE,  in  the  same  straight  line,  have 
the  same  altitude ;  therefore  (16), 

ABC  =  AC 
ABET  AE 

The  triangles  ABE,  ADE,  having  the  common  vertex  E,  and  their 
bases  AB,  AD,  in  the  same  straight  line,  have  the f same  altitude; 

therefore, 

ABE  =  AB 

ADE ~  AD' 

Multiplying  these  ratios,  we  have  (III.  14), 

ABC      AB  X  AC 
ADE  ~  AD  X  AE 


BOOK     IV.  133 


PKOPOSITION  IX.— THEOKEM. 

23.  Similar  polygons  are  to  each  other  as  the  squares  of  their  homolo 
gous  sides. 

Let  ABCDEF,  A'B' C'D'E'F',  be  two  similar  polygons;   and 
denote  their  surfaces  by  S  and 
S'i  then,  B ^ 

8 


S'      AW2 

For,  let  the  polygons  be  de 
composed  into  homologous  tri 
angles  (III.  39).     The  ratio  of  the  surfaces  of  any  pair  of  homolo 
gous  triangles,  as  ABC  and  A'B'C',  ACD  and  A'C'D',  etc.,  will  be 
the  square  of  the  ratio  of  two  homologous  sides  of  the  polygons 
(20) ;  therefore,  we  shall  have 


ABC  =    ACD    _    APE          AEF         AB 
A'B'C'  ~~  A'C'D'    "  AD'E'    '  A'E'F'  ~ 


Therefore,  by  addition  of  antecedents  and  consequents  (III.  12), 

ABC    +    ACD    +    APE   +    AEF     _S__     AB2 
A'B'C'  +  A'C'D'  +  A'D'E'  -f  A'E'F'  ~  S'~  AB72 

24.  Corollary.  The  ratio  of  the  surfaces  of  two  similar  polygons  is 
the  square  of  the  ratio  of  similitude  of  the  polygons ;  that  is,  the 
square  of  the  ratio  of  any  two  homologous  lines  of  the  polygons. 


PKOPOSITION  X.— THEOKEM. 

25.  The  square  described  upon  the  hypotenuse  of  a  right  triangle 
is  equivalent  to  the  sum  of  the  squares  described  on  the  other  two 
sides. 

12 


134 


GEOMETRY. 


Let  the  triangle  ABC  be  right  angled 
at  C;  then,  the  square  AH,  described 
upon  the  hypotenuse,  is  equal  in  area 
to  the  sum  of  the  squares  AF  and  BD, 
described  on  the  other  two  sides. 

For,  from  C  draw  CP  perpendicular 
to  AB  and  produce  it  to  meet  KH  in  L. 
Join  CK,  BO.  Since  ACF  and  ACB 
are  right  angles,  CF  and  CB  are  in 
the  same  straight  line  (I.  21) ;  and  for 
a  similar  reason  A  C  and  CD  are  in  the 
same  straight  line. 

In  the  triangles  CAK,  GAB,  we  have  AK  equal  to  AB,  being 
sides  of  the  same  square;  AC  equal  to  AO,  for  the  same  reason; 
and  the  angles  CAK,  GAB,  equal,  being  each  equal  to  the  sum 
of  the  angle  CAB  and  a  right  angle ;  therefore,  these  triangles  are 
equal  (I.  76). 

The  triangle  OJJTand  the  rectangle  AL  have  the  same  base  AK; 
and  since  the  vertex  C  is  upon  LP  produced,  they  also  have  the 
same  altitude ;  therefore,  the  triangle  CAK  is  equivalent  to  one-half 
the  rectangle  AL  (14). 

The  triangle  GAB  and  the  square  AF  have  the  same  base  AG; 
and,  since  the  vertex  B  is  upon  FC  produced,  they  also  have  the 
same  altitude;  therefore,  the  triangle  GAB  is  equivalent  to  one- 
half  the  square  AF  (14). 

But  the  triangles  CAK,  GAB,  have  been  shown  to  be  equal  ; 
therefore,  the  rectangle  AL  is  equivalent  to  the  square  AF. 

In  the  same  way,  it  is  proved  that  the  rectangle  BL  is  equivalent 
to  the  square  BD. 

Therefore,  the  square  AH,  which  is  the  sum  of  the  rectangles  AL 
and  BL,  is  equivalent  to  the  sum  of  the  squares  ^IjPand  BD. 

26.  Scholium.  This  theorem  is  ascribed  to  Pythagoras  (born  about 
600  B.  C.),  and  is  commonly  called  the  Pythagorean  Theorem.  The 
preceding  demonstration  of  it  is  that  which  was  given  by  Euclid  in 
his  Elements  (about  300  B.  C.). 

It  is  important  to  observe,  that  we  may  deduce  the  same  result 
from  the  numerical  relation  AB*  =  AC2  -f-  BC*,  already  established 
in  (III.  48).  For,  since  the  measure  of  the  area  of  a  square  is  the 


BOOK     IV.  135 

second  power  of  the  number  which  represents  its  side,  it  follows 
directly  from  this  numerical  relation  that  the  area  of  which  AB2  is 
the  measure  is  equal  to  the  sum  of  the  areas  of  which  AC2  and  BC1 
are  the  measures.  In  the  same  manner,  most  of  the  numerical  rela 
tions  demonstrated  in  the  articles  (III.  48)  to  (III.  67)  give  rise  to 
theorems  respecting  areas  by  merely  substituting,  for  a  product,  the 
area  represented  by  that  product.  This  may  be  called  a  transition 
from  the  abstract  (pure  number)  to  the  concrete  (actual  space). 

On  the  other  hand,  we  may  pass  from  the  concrete  to  the  abstract. 
For  example,  in  the  above  figure  it  has  been  proved  that  the  areas 
of  the  rectangles  AL,  BL,  are  respectively  equal  to  the  areas  of  the 
squares  AF,  BD.  But  the  rectangles,  having  the  same  altitude,  are 
to  each  other  as  their  bases  AP,  PB  ;  and  the  squares  are  to  each 
other  as  their  numerical  measures  AC*,  BC2',  hence,  we  infer  the 
numerical  relation 

AC2  :BC*  =  AP:  PB, 


which  was  otherwise  proved  in  (III.  46). 

Henceforth,  we  shall  employ  the  equation  AB2  =  AC2  -f  BC2,  as 
the  expression  of  either  one  of  the  theorems  (III.  48)  and  (IV.  25). 

27.  Corollary.  If  the  three  sides  of  a  right  triangle  be  taken  as  the 
homologous  sides  of  three  similar  polygons  constructed  upon  them,  then 
the  polygon  constructed  upon  the  hypotenuse  is  equivalent  to  the  sum  of 
the  polygons  constructed  upon  the  other  two  sides. 

For,  let  P,  Q,  R,  denote  the  areas  of  the  polygons  constructed 
upon  the  sides  AC,  BC,  and  upon  the  hypotenuse  AB,  respectively. 
Then,  the  polygons  being  similar,  we  have 


P      AC1         R      AB' 


Q       BC2         Q       BC2 

from  the  first  of  which  we  derive,  by  composition, 

P-f  Q  =  ~AC2  +  BC2  =  AB2 
Q  BC2  BC* 

which  compared  with  the  second  gives  at  once 

R  =  P  +  Q. 


136 


GEOMETRY. 


PROBLEMS    OF    CONSTRUCTION. 

PROPOSITION  XL— PROBLEM. 

28.   To  construct  a  triangle  equivalent  to  a  given  polygon. 
Let  ABCDEFbe  the  given  polygon. 
Take  any  three   consecutive  vertices,   as 
A}  B,  C,  and  draw  the  diagonal  A  C.   Through 
B  draw  BP  parallel  to  A  C  meeting  DC  pro 
duced  in  P;  join  AP. 

The  triangles  APC,  ABC,  have  the  same 
base  AC;  and  since  their  vertices,  Pand  B, 
lie  on  the  same  straight  line  BP  parallel  to  AC,  they  also  have  the 
same  altitude;  therefore  they  are  equivalent.  Therefore,  the  penta 
gon  APDEFis  equivalent  to  the  hexagon  ABCDEF.  Now,  taking 
any  three  consecutive  vertices  of  this  pentagon,  we  shall,  by  a  pre 
cisely  similar  construction,  find  a  quadrilateral  of  the  same  area ; 
and,  finally,  by  a  similar  operation  upon  the  quadrilateral,  we  shall 
find  a  triangle  of  the  same  area. 

Thus,  whatever  the  number  of  the  sides  of  the  given  polygon,  a 
series  of  successive  steps,  each  step  reducing  the  number  of  sides  by 
one,  will  give  a  series  of  polygons  of  equal  areas,  terminating  in  a 
triangle. 

PROPOSITION  XII— PROBLEM. 

29.  To  construct  a  square  equivalent  to  a  given  parallelogram  or  to  a 
given  triangle. 

1st.  Let  A  C  be  a  given  parallelogram ;   k  its     ,  B ^c 

base,  and  h  its  altitude.  \^         h\ 

Find  a  mean  proportional  x  between  h  and  k, 
by  (III.  72).  The  square  constructed  upon  x 
will  be  equivalent  to  the  parallelogram,  since 
x*  =  hX  k. 

2d.  Let  ABC  be  a  given  triangle;  a  its  base 
and  h  its  altitude. 

Find  a  mean  proportional  x  between  a  and 
i  h ;  the  square  constructed  upon  x  will  be 
equivalent  to  the  triangle,  since  x2  =  a  X  %h 


BOOKIV.  137 

30.  Scholium.  By  means  of  this  problem  and   the  preceding,  a 
square  can  be  found  equivalent  to  any  given  polygon. 


PROPOSITION  XIII.—  PROBLEM. 

31.   To  construct  a  square  equivalent  to  the  sum  of  two  or  more  given 
squares,  or  to  the  difference  of  two  given  squares. 

1st.  Let  m,  nt  PJ  q,  be  the  sides  of  given  squares. 

Draw  AB  —  m,  and  BC  =  n,  perpendicular  to 
each  other  at  B]  join  AC.  Then  (25),  AC*  = 
m2  -f  n\ 

Draw  CD  =  p,  perpendicular  to  AC,  and  join 
AD. 


Draw  DE  =  q  perpendicular  to  AD,  and  join 
AE.  Then,  AE'2  =  AD*  +  q2  =  m2  +  n2  + 
p'2  -f-  <?2;  therefore,  the  square  constructed  upon 
AE  will  be  equivalent  to  the  sum  of  the  squares 
constructed  upon  m,  n,  p,  q. 

In  this  manner  may  the  areas  of  any  number  of  given  squares  be 
added. 

2d.  Construct  a  right  angle  ABC,  and  lay  off 
BA  —  n.  With  the  centre  A  and  a  radius  =  m, 
describe  an  arc  cutting  BC  in  C.  Then  BCZ  = 

XJ  #"  ^ 

AC  2  —  AB2  =  m2  —  nz>,  therefore,  the  square  con 

structed  upon  BC  will  be  equivalent  to  the  difference  of  the  squares 

constructed  upon  m  and  n. 

32.  Scholium  I.  By  means  of  this  problem,  together  with  the  pre 
ceding  ones,  a  square  can  be  found  equivalent  to  the  sum  of  any 
number  of  given  polygons  ;  or  to  the  difference  of  any  two  given 
polygons. 

33.  /Scholium  II.  If  m,  n,  p,  q,  in  the  preceding  problem  are  ho 
mologous  sides  of  given  similar  polygons,  the  line  AE  in  the  first 
figure  is  the  homologous  side  of  a  similar  polygon  equivalent  to  the 
sum  of  the  given  polygons  (27). 

And  the  line  BC,  in  the  second  figure,  is  the  homologous  side  of  a 
similar  polygon,  equivalent  to  the  difference  of  two  given  similar 
polygons. 


138 


GEOMETRY. 


One  side  of  a  polygon,  similar  to  a  given  polygon,  being  known, 
the  polygon  may  be  constructed  by  (III.  80). 


PKOPOSITION  XIV.— PEOBLEM. 


34.    Upon  a  given  straight 
a  given  rectangle. 

Let  lc'  be  the  given  straight  line,  and  A  C the 
given  rectangle  whose  base  is  k  and  altitude  h. 

Find  a  fourth  proportional  h',  to  Jcf,  k  and  h, 
by  (III.  70).  Then,  the  rectangle  constructed 
upon  the  base  lc'  with  the  altitude  h'  is  equiva 
lent  to  AC-9  for,  by  the  construction,  kf  :  k  = 
h  :  h',  whence,  k'  X  h'  =  k  X  h  (7). 


to  construct  a  rectangle  equivalent  to 


PKOPOSITION  XV.— PROBLEM. 

35.   To  construct  a  rectangle,  having  given  its  area  and  the  sum  of 
two  adjacent  sides. 

Let  MN  be  equal  to  the  given  sum  of  the 
adjacent  sides  of  the  required  rectangle ;  and 
let  the  given  area  be  that  of  the  square  whose 
side  is  AB. 

Upon  MN  as  a  diameter  describe  a  semi 
circle.  At  M  erect  MP  =  AS  perpendicular 
to  MN,  and  draw  PQ  parallel  to  MN,  intersecting  the  circumference 
in  Q.  From  Q  let  fall  QE  perpendicular  to  MN;  then,  ME  and 
EN  are  the  base  and  altitude  of  the  required  rectangle.  For,  by 
(III.  47),  MR  XEN=  §7P  =  PM2  = 


R  N 


PROPOSITION  XVI.— PROBLEM. 

36.   To  construct  a  rectangle,  having  given  its  area  and  the  difference 
of  two  adjacent  sides. 


BOOK    IV. 


139 


Let  MN  be  equal  to  the  given  difference  of 
the  adjacent  sides  of  the  required  rectangle; 
and  let  the  given  area  be  that  of  the  square 
described  on  AB. 

Upon  MN  as  a  diameter  describe  a  circle. 
At  M  draw  the  tangent  MP  =  AB,  and  from 
P,  draw  the  secant  PQR  through  the  centre  of 
the  circle ;  then,  PR  and  PQ  are  the  base  and 
altitude  of  the  required  rectangle.  For,  by  (III.  59),  PR  X  PQ  = 
PM2  =  AB2,  and  the  difference  of  PR  and  PQ  is  QR  =  MN. 


PKOPOSITION  XVII.— PKOBLEM. 

37.   To  find  two  straight  lines  in  the  ratio  of  the  areas  of  two  given 
polygons. 

Let  squares  be  found  equal  in  area  to  the 
given  polygons,  respectively  (30).     Upon   the 
sides  of  the  right  angle  ACS,  take  CA  and  CB 
equal  to  the  sides  of  these  squares,  join  AB  and 
let  fall   CD  perpendicular  to  AB.     Then,  by  (III.  46),  we  have 
AD  :  DB  =  CA2  :  CB2',  therefore,  AD,  DB,  are  in  the  ratio  of 
the  areas  of  the  given  polygons. 


PEOPOSITION  XVIII.— PKOBLEM. 

38.  To  find  a  square  which  shall  be  to  a  given  square  in  the  ratio  of 
two  given  straight  lines. 

Let  AB2  be  the  given  square,  and  M  :  N 
the  given  ratio. 

Upon  an  indefinite  straight  line  CL,  lay 
off  CD  =  M,  DE  =  N-,  upon  CE  as  a 
diameter  describe  a  semicircle ;  at  D  erect 
the  perpendicular  DF  cutting  the  circum 
ference  in  F;  join  FC,  FE;  lay  off  FH  =  AB,  and  through  H 
draw  HG  parallel  to  EC;  then,  FG  is  the  side  of  the  required 
square.  For,  by  (III.  15),  we  have 


140  GEOMETRY. 

whence  (III.  13), 


=  FC*  :  FE\ 
Also,  by  (III.  46), 

FC2  :  FE2  =  CD  :  DE=M:  N. 
Hence, 

2  =  M:N. 


But  FH  =  AB,  therefore  the  square  constructed  upon  FG  is  to  the 
square  upon  AB  in  the  ratio  M :  N. 

PROPOSITION  XIX.— PROBLEM. 

39.  To  construct  a  polygon  similar  to  a  given  polygon  and  whose  area 
shall  be  in  a  given  ratio  to  that  of  the  given  polygon. 

Let  P  be  the  given  polygon,  and  let  a  be  one  of 
its  sides ;  let  M :  N  be  the  given  ratio. 

Find,  by  the  preceding  problem,  the  side  a'  of  a 
square  which  shall  be  to  a2  in  the  ratio  M  :  N ; 
upon  a',  as  a  homologous  side  to  a,  construct  the 
polygon  P'  similar  to  P  (III.  80)  ;  this  will  be  the 
polygon  required. 

For,  the  polygons  being  similar,  their  areas  are 
in  the  ratio  a'2 :  a2,  or  M  :  N,  as  required. 

PROPOSITION  XX.— PROBLEM. 

40.  To  construct  a  polygon  similar  to  a  given  polygon  P  and  equiva 
lent  to  a  given  polygon  Q.  > ' 

Find  M  and  N,  the  sides  of  squares 
respectively  equal  in  area  to  P  and  Q, 
(30). 

Let  a  be  any  side  of  P,  and  find  a 
fourth  proportional  a'  to  M,  N  and  a: 
upon  a',  as  a  homologous  side  to  a,  con 
struct  the  polygon  P'  similar  to  P;  this 
will  be  the  required  polygon.  For,  by 

construction, 

M       a 


BOOK    IV.  141 

therefore,  taking  the  letters  P,  Q  and  P',  to  denote  the  areas  of  the 
polygons, 

~q  =  W2^^ 

but,  the  polygons  P  and  P'  being  similar,  we  have,  by  (23), 

Z  —  £!. 
P'~  a'2' 

and  comparing  these  equations,  we  have  P'  =  Q. 

Therefore,  the  polygon  P'  is  similar  to  the  polygon  P  and  equiva 
lent  to  the  polygon  Q,  as  required. 


BOOK   V. 

REGULAR    POLYGONS.      MEASUREMENT    OF    THE    CIRCLE. 
MAXIMA  AND  MINIMA    OF  PLANE  FIGURES. 

REGULAR  POLYGONS. 

1.  DEFINITION.  A  regular  polygon  is  a  polygon  which  is  at  once 
equilateral  and  equiangular. 

The  equilateral  triangle  and  the  square  are  simple  examples  of 
regular  polygons.  The  following  theorem  establishes  the  possibility 
of  regular  polygons  of  any  number  of  sides. 


PEOPOSITION  I.— THEOREM. 

2.  If  the  circumference  of  a  circle  be  divided  into  any  number  of 
equal  parts,  the  chords  joining  the  successive  points  of  division  form  a 
regular  polygon  inscribed  in  the  circle;  and  the  tangents  drawn  at  the 
points  of  division  form  a  regular  polygon  circumscribed  about  the  circle. 

Let  the  circumference  be  divided  into  the 
equal  arcs  AB,  EC,  CD,  etc. ;  then,  1st,  draw 
ing  the  chords  AB,  BC,  CD,  etc.,  ABCD,  etc., 
is  a  regular  inscribed  polygon.  For,  its  sides 
are  equal,  being  chords  of  equal  arcs;  and 
its  angles  are  equal,  being  inscribed  in  equal 
segments. 

2d.  Drawing  tangents  at  A,  B,  C,  etc.,  the 
polygon  GHK,  etc.,  is  a  regular  circumscribed 

polygon.  For,  in  the  triangles  AGB,  BHC,  CKD,  etc.,  we  have 
AB  =  BC  =  CD,  etc.,  and  the  angles  GAB,  GBA,  HBC,  HCB, 
etc.,  are  equal,  since  each  is  formed  by  a  tangent  and  chord  and  is 
measured  by  half  of  one  of  the  equal  parts  of  the  circumference 
142 


BOOK     V.  143 

(II.  62)  ;  therefore,  these  triangles  are  all  isosceles  and  equal  to  each 
other.  Hence,  we  have  the  angles  G  =  H  =  K,  etc.,  and  A  G  = 
GB  =  BH  =  HC  =  CK,  etc.,  from  which,  by  the  addition  of 
equals,  it  follows  that  GH  =  UK,  etc. 

3.  Corollary  I.  Hence,  if  an  inscribed  polygon  is  given,  a  circum 
scribed  polygon  of  the  same  number  of  sides  can  be  formed  by 
drawing  tangents  at  the  vertices  of  the  given  polygon.     And  if  a 
circumscribed  polygon  is  given,  an  inscribed  polygon  of  the  same 
number  of  sides  can  be  formed  by  joining  the  points  at  which  the 
sides  of  the  given  polygon  touch  the  circle. 

It  is  often  preferable,  however,  to  obtain  the  circumscribed  polygon 
from  the  inscribed,  and  reciprocally,  by  the  following  methods : 

1st.  Let  ABCD....  be  a  given  inscribed  polygon.  Bisect  the 
arcs  AB,  BC,  CD,  etc.,  in  the  points  E,  F, 
G,  etc.,  and  draw  tangents,  A'B',B'C', 
C'D',  etc.,  at  these  points ;  then,  since  the 
arcs  EF,  FG,  etc.,  are  equal,  the  polygon 
A'B' C'D' ....  is,  by  the  preceding  propo 
sition,  a  regular  circumscribed  polygon  of 
the  same  number  of  sides  as  ABCD .... 
Since  the  radius  OE  is  perpendicular  to 

AB  (II.  16)  as  well  as  to  A'B1,  the  sides  A'B',  AB,  are  parallel; 
and,  for  the  same  reason,  all  the  sides  of  A'B' C'D' ....  are  parallel 
to  the  sides  of  ABCD . . .  .  respectively.  Moreover,  the  radii  OA, 
OB,  OC,  etc.,  when  produced,  pass  through  the  vertices  A',B^C',  etc. ; 
for  since  B'E  =  B'F,  the  point  B'  must  lie  on  the  line  OB  which 
bisects  the  angle  EOF  (I.  127). 

2d.  If  the  circumscribed  polygon  A'B' C'D' is  given,  we  have 

only  to  draw  OA't  OB',  OC',  etc.,  intersecting  the  circumference  in 
A,  B,  C,  etc.,  and  then  to  join  AB,  BC,  CD,  etc.,  to  obtain  the  in 
scribed  polygon  of  the  same  number  of  sides. 

4.  Corollary  II.  If  the  chords  AE,  EB,  BF,  FC,  etc.,  be  drawn, 
a  regular  inscribed  polygon  will  be  formed  of  double  the  number  of 
sides  of  ABCD 

If  tangents  are  drawn  at  A,  B,  C,  etc.,  intersecting  the  tangents 
A'B',  B'C',  C'D',  etc.,  a  regular  circumscribed  polygon  will  be 
formed  of  double  the  number  of  sides  of  A'B' C'D'. . . . 

It  is  evident  that  the  area  of  an  inscribed  polygon  is  less  than 


144  GEOMETRY. 

that  of  the  inscribed  polygon  of  double  the  number  of  sides ;  and 
the  area  of  a  circumscribed  polygon  is  greater  than  that  of  the  cir 
cumscribed  polygon  of  double  the  number  of  sides. 


PROPOSITION  II:— THEOREM. 

5.  A  circle  may  be  circumscribed  about  any  regular  polygon ;  and  a 
circle  may  aho  be  inscribed  in  it. 

Let  ABCD ...  be  a  regular  polygon ;  then, 

1st.  A  circle  may  be  circumscribed  about 
it.  For,  describe  a  circumference  passing 
through  three  consecutive  vertices  A,  B,  G 
(II.  88) ;  let  0  be  its  centre,  draw  OH  per 
pendicular  to  B  C  and  bisecting  it  at  H,  and 
join  OA,  OD.  Conceive  the  quadrilateral 

A  OHB  to  be  revolved  upon  the  line  OH  (i.  e.,  folded  over),  until 
HB  falls  upon  its  equal  HC.  The  polygon  being  regular,  the  angle 
HBA  =  HCD,  and  the  side  BA  =  CD;  therefore  the  side  BA  will 
take  the  direction  of  CD  and  the  point  A  will  fall  upon  D.  Hence 
OD  =  OA,  and  the  circumference  described  with  the  radius  OA 
and  passing  through  the  three  consecutive  vertices  A,  B,  C,  also 
passes  through  the  fourth  vertex  D.  It  follows  that  the  circumfer 
ence  which  passes  through  the  three  vertices  B,  C,  D,  also  passes 
through  the  next  vertex  E,  and  thus  through  all  the  vertices  of  the 
polygon.  The  circle  is  therefore  circumscribed  about  the  polygon. 

2d.  A  circle  may  be  inscribed  in  it.  For,  the  sides  <of  the  polygon 
being  equal  chords  of  the  circumscribed  circle,  are  equally  distant 
from  the  centre  ;  therefore,  a  circle  described  with  the  centre  0  and 
the  radius  OH  will  touch  all  the  sides,  and  will  consequently  be  in 
scribed  in  the  polygon. 

6.  Definitions.  The  centre  of  a  regular  polygon  is  the  common  cen 
tre,  0,  of  the  circumscribed  and  inscribed  circles. 

The  radius  of  a  regular  polygon  is  the  radius,  OA,  of  the  circum 
scribed  circle. 

The  apothem  is  the  radius,  OH,  of  the  inscribed  circle. 

The  angle  at  the  centre  is  the  angle,  A  OB,  formed  by  radii  drawn 
to  the  extremities  of  any  side. 


BOOK      V. 


145 


7.  The  angle  at  the  centre  is  equal  to  four  right  angles  divided  by 
the  number  of  sides  of  the  polygon. 

8.  Since  the  angle  ABC  is  equal  to  twice  ABO,  or  to  ABO  -f- 
BAO,  it  follows  that  the  angle  ABC  of  the  polygon  is  the  supple 
ment  of  the  angle  at  the  centre  (I.  68). 


PROPOSITION  III.— THEOREM. 

9.  Regular  polygons  of  the  same  number  of  sides  are  similar. 
Let  ABODE,  A'B'C'D'E',  be 

regular  polygons  of  the  same  num 
ber  of  sides ;  then,  they  are  similar. 

For,  1st,  they  are  mutually  equi 
angular,  since  the  magnitude  of 
an  angle  of  either  polygon  de 
pends  only  on  the  number  of  the 
sides  (7  and  8),  which  is  the  same  in  both. 

2d.   The    homologous    sides    are    proportional,   since    the    ratio 
AB  :  A'B'  is  the  same  as  the  ratio  BC :  B'C',  or  CD  :  C'D',  etc. 

Therefore  the  polygons  fulfill  the  two  conditions  of  similarity. 

10.  Corollary.   The  perimeters  of  regular  polygons  of  the  same  num 
ber  of  sides  are  to  each  other  as  the  radii  of  the  circumscribed  circles, 
or  as  the  radii  of  the  inscribed  circles ;  and  their  areas  are  to  each  other 
as  the  squares  of  these  radii.     For,  these  radii  are  homologous  lines 
of  the  similar  polygons  (III.  43),  (IV.  24). 


PEOPOSITION  IV.— PROBLEM. 
11.   To  inscribe  a  square  in  a  given  circle. 

Draw  any  two  diameters  AC,  BD,  perpen 
dicular  to  each  other,  and  join  their  extremities 
by  the  chords  AB,  BC,  CD,  DA  ;  then,  ABCD 
is  an  inscribed  square  (II.  12),  (II.  59), 


12.   Corollary.  To  circumscribe  a  square  about  the  circle,  draw 
tangents  at  the  extremities  of  two  perpendicular  diameters  A  C,  BDf 

13  K 


146  GEOMETRY. 

13.  Scholium.  In  the  right  triangle  ABO,  we  have  JLB2  = 
OT  +  OB1  --=  20A2,  whence  AB  =  OA.  i/2,  by  which  the  side  of 
the  inscribed  square  can  be  computed,  the  radius  being  given. 


PKOPOSITION  V.— PEOBLEM. 

14.  To  inscribe  a  regular  hexagon  in  a  given  circle. 
Suppose    the    problem     solved,    and     let 

ABCDEF\)Q  a  regular  inscribed  hexagon. 

Join  .RE and  AD;  since  the  arcs  AB,  BC, 
CD,  etc.,  are  equal,  the  lines  BE,  AD,  bisect 
the  circumference  and  are  diameters  inter 
secting  in  the  centre  0.  The  inscribed  angle 
ABO  is  measured  by  one-half  the  arc  AFE, 
that  is,  by  AF,  or  one  of  the  equal  divisions  of  the  circumference  ; 
the  angle  A  OB  at  the  centre  is  also  measured  by  one  division,  that 
is,  by  AB-,  and  the  angle  BAO  ==  ABO;  therefore  the  triangle 
ABO  is  equiangular,  and  AB  =  OA.  Therefore  the  side  of  the 
inscribed  regular  hexagon  is  equal  to  the  radius  of  the  circle. 

Consequently,  to  inscribe  a  regular  hexagon,  apply  the  radius  six 
times  as  a  chord. 

15.  Corollary.  To  inscribe  an  equilateral  triangle,  A  CE,  join  the 
alternate  vertices  of  the  regular  hexagon. 

16.  Scholium.    In    the   right    triangle   ACDt  we  have  AC2  = 
AD2  —  DC2  =  (2 A O)2  —  JO2  =  3Z02;  whence,  AC  =  A  0.1/3, 
by  which  the  side  of  the  inscribed  equilateral  triangle  can  be  com 
puted,  the  radius  being  given. 

The  apothem,  OH,  of  the  inscribed  equilateral  triangle  is  equal  to 
one-half  the  radius  OB;  for  the  figure  AOCB  is  a  rhombus  and  its 
diagonals  bisect  each  other  at  right  angles  (I.  110). 

The  apothem  of  the  inscribed  regular  hexagon  is  equal  to  one-half 

the  side  of  the  inscribed  equilateral  triangle,  that  is,  to 1/3  ;  for 

the  perpendicular  from  0  upon  AB  is  equal  to  the  perpendicular 
from  A  upon  OB,  that  is,  to  AH. 

The  angle  at  the  centre  of  the  regular  inscribed  hexagon  is  ^  of 
4  right  angles,  that  is,  -f  of  one  right  angle  =  60°. 


BOOK     V. 


147 


The  angle  of  the  hexagon,  or  ABC,  is  f  of  a  right  angle  =  120°. 
The  angle  at  the  centre  of  the  inscribed  equilateral  triangle  is  f  of 
one  right  angle  =  120°. 


PROPOSITION  VI.— PEOBLEM. 

17.   To  inscribe  a  regular  decagon  in  a  given  circle. 

Suppose  the  problem  solved,  and  let 
ABC . ...  X,  be  a  regular  inscribed  decagon. 

Join  AF,  BG;  since  each  of  these  lines  bi 
sects  the  circumference,  they  are  diameters  and 
intersect  in  the  centre  0.  Draw  BK  intersect 
ing  OA  in  M. 

The  angle  AMB  is  measured  by  half  the 
sum  of  the  arcs  KF  and  AB  (II.  64),  that  is, 

by  two  divisions  of  the  circumference ;  the  inscribed  angle  MAB  is 
measured  by  half  the  arc  BF,  that  is,  also  by  two  divisions ;  there 
fore  AMB  is  an  isosceles  triangle,  and  MB  =  AB. 

Again,  the  inscribed  angle  MBO  is  measured  by  half  the  arc  KG, 
that  is,  by  one  division,  and  the  angle  MOB  at  the  centre  has  the 
same  measure ;  therefore  OMB  is  an  isosceles  triangle,  and  OM  = 
MB  =  AB. 

The  inscribed  angle  MBA,  being  measured  by  half  the  arc  AK, 
that  is,  by  one  division,  is  equal  to  the  angle  AOB.  Therefore  the 
isosceles  triangles  AMB  and  AOB  are  mutually  equiangular  and 
similar,  and  give  the  proportion 


whence 


OA  :  AB  =  AB  :  AM, 
OA  X  AM=  AB*  =  OF2; 


that  is,  the  radius  OA  is  divided  in  extreme  and  mean  ratio  at  M 
(III.  73)  ;  and  the  greater  segment  OM  is  equal  to  the  side  AB  of 
the  inscribed  regular  decagon. 

Consequently,  to  inscribe  a  regular  decagon,  divide  the  radius  in 
extreme  and  mean  ratio  (III.  74),  and  apply  the  greater  segment  ten 
times  as  a  chord. 


148 


GEOMETRY. 


18.  Corollary.  To  inscribe  a  regular  penta 
gon,  A  CEGK,  join  the  alternate  vertices  of  the 
regular  inscribed  decagon. 

19.  Scholium.  By  (III.  75),  we  have 


AB  = 


by  which  the  side  of  the  regular  decagon  may  be  computed  from  the 
radius. 

The  angle  at  the  centre  of  the  regular  decagon  is  f  of  one  right 
angle  =  36° ;  the  angle  at  the  centre  of  the  regular  pentagon  is  f 
of  one  right  angle  =  72°. 

The  angle  ABC  of  the  regular  decagon  is  f  of  one  right  angle  = 
144° ;  the  angle  A  CE  of  the  regular  pentagon  is  f  of  one  right 
angle  =  108°. 


PROPOSITION  VII.-PROBLEM. 

20.  To  inscribe  a  regular  pentedecagon  in  a  given  circle. 
Suppose  AB  is  the  side  of  a  regu 
lar  inscribed  pentedecagon,  or  that 

the  arc  AB  is  ^  of  the  circumfer- 

B 

ence. 

Now  the  fraction  TV  —  i  —  TV  '•>  therefore  the  arc  A B  is  the  dif 
ference  between  ^  and  y1^  of  the  circumference.  Hence,  if  we 
inscribe  the  chord  AC  equal  to  the  side  of  the  regular  inscribed 
hexagon,  and  then  CB  equal  to  that  of  the  regular  inscribed  decagon, 
the  chord  AB  will  be  the  side  of  the  regular  inscribed  pentedecagon 
required. 

21.  Scholium.  Any  regular  inscribed  polygon  being  given,  a  regu 
lar  inscribed  polygon  of  double  the  number  of  sides  can  be  formed 
by  bisecting  the  arcs  subtended  by  its  sides  and  drawing  the  chords 
of  the  semi-arcs  (4).     Also,  any  regular  inscribed  polygon   being 
given,  a  regular  circumscribed  polygon  of  the  same  number  of  sides 
can  be  formed  (3).     Therefore,  by  means  of  the  ins9ribed  square,  we 
can  inscribe  and  circumscribe,  successively,  regular  polygons  of  8, 
16,  32,  etc.,  sides ;  by  means  of  the  hexagon,  those  of  12,  24,  48,  etc., 
sides ;  by  means  of  the  decagon,  those  of  20,  40,  80,  etc.,  sides  ;  and, 


BOOK    V. 


149 


finally,  by  means  of  the  pentedecagon,  those  of  30,  60,  120,  etc., 
sides. 

Until  the  beginning  of  the  present  century,  it  was  supposed  that 
these  were  the  only  polygons  that  could  be  constructed  by  elementary 
geometry,  that  is,  by  the  use  of  the  straight  line  and  circle  only. 
GAUSS,  however,  in  his  Disquisitiones  Arithmeticce,  Lipsice,  1801, 
proved  that  it  is  possible,  by  the  use  of  the  straight  line  and  circle 
only,  to  construct  regular  polygons  of  17  sides,  of  257  sides,  and  in 
general  of  any  number  of  sides  which  can  be  expressed  by  2n  -f-  1, 
n  being  an  integer,  provided  that  2n  -f-  1  is  a  prime  number. 


PEOPOSITION  VIII.— THEOEEM. 

22.  The  area  of  a  regular  polygon  is  equal  to  half  the  product  of  its 
perimeter  and  apothem. 

For,  straight  lines  drawn  from  the  centre  to  the  vertices  of  the 
polygon  divide  it  into  equal  triangles  whose  bases  are  the  sides  of 
the  polygon  and  whose  common  altitude  is  the  apothem.  The  area 
of  one  of  these  triangles  is  equal  to  half  the  product  of  its  base  and 
altitude  (IV.  13) ;  therefore,  the  sum  of  their  areas,  or  the  area 
of  the  polygon,  is  half  the  product  of  the  sum  of  the  bases  by  the 
common  altitude,  that  is,  half  the  product  of  the  perimeter  and 
apothem. 

PEOPOSITION  IX.— THEOEEM. 

23.  The  area  of  a  regular  inscribed  dodecagon  is  equal  to  three  times 
the  squgre  of  the  radius. 

Let  AB,  BC,  CD,  DE,  be  four  consecu 
tive  sides  of  a  regular  inscribed  dodeca 
gon,  and  draw  the  radii  OA,  OE;  then, 
the  figure  0 ABODE  is  pne-third  of  the 
dodecagon,  and  we  have  only  to  prove 
that  the  area  of  this  figure  is  equal  to 
the  square  of  the  radius. 

Draw  the  radius  OD ;  at  A  and  D  draw 
the  tangents  AF  and  GDF  meeting  in  F; 

join  AC  and  CE,  and  let  AC  and  OEloe  produced  to  meet  the  tan- 
is  * 


TT 


150 


GEOMETRY. 


n 


gent  OF'm  H  and  Q.     The  arc  AD,  containing  three  of  the  sides 
of  the  dodecagon,  is  one  fourth  of  the  cir 
cumference  ;  therefore  the  angle  A  OD  is 
a  right  angle,  and   OF  is  a  square  de 
scribed  on  the  radius. 

Since  A  C  and  CE  are  sides  of  the  regu 
lar  inscribed  hexagon,  each  is  equal  to 
the  radius ;  therefore  OA  CE  is  a  paral 
lelogram.  Hence  also  GOAHaud  GECH 
are  parallelograms. 

The  triangles  DEC  and  BCA  are  equal  (I.  80).  The  area  of  the 
triangle  DEC  is  one-half  that  of  the  parallelogram  EH  (IV.  14) ; 
therefore  the  two  triangles  DEC  and  BCA  are  together  equivalent 
to  the  parallelogram  EH.  Adding  the  parallelogram  OC  to  these 
equals,  we  have  the  figure  OABCDE  equivalent  to  the  parallelogram 
OH.  But  the  parallelogram  OH  is  equivalent  to  the  square  OF 
(IV.  11);  therefore  the  figure  OABCDE,  or  one-third  the  dodecagon, 
is  equivalent  to  the  square  OF,  that  is,  to  the  square  of  the  radius! 
Therefore,  the  area  of  the  whole  dodecagon  is  equal  to  three  times 
the  square  of  the  radius. 

24.  Scholium.  The  area  of  the  circumscribed  square  is  evidently 
equal  to  four  times  the  square  of  the  radius.  The  area  of  the  circle 
is  greater  than  that  of  the  inscribed  regular  dodecagon,  and  less  than 
that  of  the  circumscribed  square;  therefore,  if  the  square  of  the 
radius  is  taken  as  the  unit  of  surface,  the  area  of  a  circle  is  greater 
than  3  and  less  than  4. 


PROPOSITION  X.— PROBLEM. 

25.  Given  the  perimeters  of  a  regular  inscribed  and  a  similar  cir 
cumscribed  polygon,  to  compute  the  perimeters  of  the  regular  inscribed 
and  circumscribed  polygons  of  double  the  number  of  sides. 

Let  AB  be  a  side  of  the  given  inscribed  polygon,  CD  a  side  of  the 
similar  circumscribed  polygon,  tangent  to  the  arc  AB  at  its  middle 
point  E.  Join  AE,  and  at  A  and  B  draw  the  tangents  AF  and 
BG\  then  AE  is  a  side  of  the  regular  inscribed  polygon  of  double 


BOOK    V.  151 

the  number  of  sides,  and  FG  is  a  side      c          F        E        G         D 
of  the  circumscribed  polygon  of  double       ' 
the  number  of  sides  (4).  ~/^T 

Denote  the  perimeters  of  the  given 
inscribed  and  circumscribed  polygons 
by  p  and  P,  respectively ;  and  the  pe 
rimeters  of  the  required  inscribed  and  Nj[/ 
circumscribed  polygons  of  double  the 
number  of  sides  by  p'  and  P',  respectively. 

Since  OC  is  the  radius  of  the  circle  circumscribed  about  the  poly 
gon  whose  perimeter  is  P,  we  have  (10), 

P_  OC       OC 
p==~OA°TOE' 

and  since  OF  bisects  the  angle  COE,  we  have  (III.  21), 

OC  _  OF. 

OE~  FE' 

therefore, 

P__CF 

P~  FE 
whence,  by  composition, 

p  +  P  __  CF  +  FE       CE 
2p  2FE       ~!?G 

Now  FG  is  a  side  of  the  polygon  whose  perimeter  is  P',  and  is  con 
tained  as  many  times  in  P'  as  CE  is  contained  in  P,  hence  (III.  9), 

CE_P_ 
FG~Pf' 

and  therefore, 

P  +  p  _P 
2p       ~  P'' 
whence 


~P  +  P  [i] 

Again,  the  right  triangles  AEH  and  EFNare  similar,  since  their 


152 


GEOMETRY. 


acute  angles  EAHand  FEN  are  equal, 

and  give 

AH^EN 

AE  ~=  EF 


7\ 


Since  J.H  and  J.-E  are  contained  the 
same  number  of  times  in  p  and  p  ',  re 
spectively,  we  have 

AH       p 


and  since  .E2V  and  .EF  are  contained  the  same  number  of  times  in 
p'  and  P',  respectively,  we  have 


therefore,  we  have 


=_ 

EF  ~  P' ' 

P_  ==  £' 
Pf       P'' 


whence 


[2] 


Therefore,  from  the  given  perimeters  p  and  P,  we  compute  P'  by 
the  equation  [1],  and  then  with  p  and  P'  we  compute  p'  by  the 
equation  [2]. 

26.  Definition.  Two  polygons  are  isoperimetrie  when  their  perime 
ters  are  equal. 

PROPOSITION  XI.— PROBLEM. 

27.  Given  the  radius  and  apothem  of  a  regular  polygon,  to  compute 


num- 


the  radius  and  apothem  of  the  isoperimetrie  polygon  of  double  the 

ber  of  sides. 

Let  AB  be  a  side  of  the  given  regular  polygon,  0  the  centre  c 
this  polygon,  OA  its  radius,  OD  its  apothem. 
Produce  DO  to  meet  the  circumference  of  the 
circumscribed  circle  in  0' ';  join  O'A,  0'B\ 
let  fall  OA'  perpendicular  to  O'A,  and 
through  A'  draw  A'B'  parallel  to  AB. 

Since  the  new  polygon  is  to  have  twice  as 
many  sides  as  the  given  polygon,  the  angle  at 
its  centre  must  be  one-half  the  angle  A  OB ; 


BOOK     V.  153 

therefore  the  angle  AO'B,  which  is  equal  to  one-half  of  AOB 
(II.  57),  is  equal  to  the  angle  at  the  centre  of  the  new  polygon. 

Since  the  perimeter  of  the  new  polygon  is  to  be  equal  to  that  of 
the  given  polygon,  but  is  to  be  divided  into  twice  as  many  sides, 
each  of  its  sides  must  be  equal  to  one-half  of  AB ;  therefore  AB', 
which  is  equal  to  one-half  of  AB  (I.  121),  is  a  side  of  the  new 
polygon  ;  O'A  is  its  radius,  and  O'D'  its  apothem. 

If,  then,  we  denote  the  given  radius  OA  by  R,  and  the  given 
apothem  OD  by  r,  the  required  radius  O'A  by  Rr,  and  the  apothem 
O'D'  by  r',  we  have 

/       O'D       00'  +  OD 

OD  =  ~T         ~T 

or 

r'=^p.  PI 

In  the  right  triangle  OA'0't  we  have  (III.  44), 


0'A=  00'  X  O'D', 
or 

R'  =  VR~X?;  [2] 

therefore,  r'  is  an  arithmetic  mean  between  R  and  r,  and  R'  is  a 
geometric  mean  between  R  and  r'. 

MEASUREMENT    OF    THE    CIKCLE. 

The  principle  which  we  employed  in  the  comparison  of  incommen 
surable  ratios  (II.  49)  is  fundamentally  the  same  as  that  which  we 
are  about  to  apply  to  the  measurement  of  the  circle,  but  we  shall 
now  state  it  in  a  much  more  general  form,  better  adapted  for  subse 
quent  application. 

28.  Definitions.  I.  A  variable  quantity,  or  simply,  a  variable,  is  a 
quantity  which  has  different  successive  values. 

II.  When  the  successive  values  of  a  variable,  under  the  conditions 
imposed  upon  it,  approach  more  and  more  nearly  to  the  value  of 
some  fixed  or  constant  quantity,  so  that  the  difference  between  the 
variable  and  the  constant  may  become  less  than  any  assigned  quan 
tity,  without  becoming  zero,  the  variable  is  said  to  approach  indefi- 


154  GEOMETRY. 

nitely  to  the  constant  ;  and  the  constant  is  called  the  limit  of  the 
variable. 

Or,  more  briefly,  the  limit  of  a  variable  is  a  constant  quantity  to 
which  the  variable,  under  the  conditions  imposed  upon  it,  approaches 
indefinitely. 

As  an  example,  illustrating  these  definitions,  let  a  point  be  re 
quired  to  move  from  A  to  B  under  the  fol 

lowing  conditions  :  it  shall  first  move  over  j  __  .  i  ^  i 
one-half  of  AB,  that  is  to  C;  then  over  A  B 

one-half  of  CB,  to  C'  ;  then  over  one-half  of  C'B,  to  C"  ;  and  so 
on  indefinitely  ;  then  the  distance  of  the  point  from  A  is  a  variable, 
and  this  variable  approaches  indefinitely  to  the  constant  AB,  as  its 
limit,  without  ever  reaching  it. 

As  a  second  example,  let  A  denote  the  angle  of  any  regular  poly 
gon,  and  n  the  number  of  sides  of  the  polygon  ;  then,  a  right  angle 
being  taken  as  the  unit,  we  have  (8), 


The  value  of  J.  is  a  variable  depending  upon  n  ;  and  since  n  may  be 

4 
taken  so  great  that  —  shall  be  less  than  any  assigned  quantity  how 

ever  small,  the  value  of  A  approaches  to  two  right  angles  as  its  limit, 
but  evidently  never  reaches  that  limit. 

29.  PRINCIPLE  OF  LIMITS.     Theorem.    If  two  variable  quantities 
are  always  equal  to  each  other  and  each  approaches  to  a  limit,  the  two 
limits  are  necessarily  equal. 

For,  two  variables  always  equal  to  each  other  present  in  fact  but 
one  value,  and  it  is  evidently  impossible  that  one  variable  value 
shall  at  the  same  time  approach  indefinitely  to  two  unequal  limits. 

30.  Theorem.   The  limit  of  the  product  of  two  variables  is  the  pro 
duct  of  their  limits.     Thus,  if  x  approaches  indefinitely  to  the  limit  a, 
and  y  approaches  indefinitely  to  the  limit  b,  the  product  xy  must 
approach  indefinitely  to  the  product  ab  ;  that  is,  the  limit  of  the  pro 
duct  xy  is  the  product  ab  of  the  limits  of  x  and  y. 

31.  Theorem.   If  two   variables  are  in  a  constant  ratio  and  each 
approaches  to  a  limit,  these  limits  are  in  the  same  constant  ratio. 

Let  x  and  y  be  two  variables  in  the  constant  ratio  m,  that  is,  let 


BOOK    V.  155 

x  =  my;  and  let  their  limits  be  a  and  b  respectively,  Since  y  ap 
proaches  indefinitely  to  b,  my  approaches  indefinitely  to  mb ;  there 
fore  we  have  x  and  my,  two  variables,  always  equal  to  each  other, 
whose  limits  are  a  and  mb,  respectively,  whence,  by  (29),  a  =  mb ; 
that  is,  a  and  b  are  in  the  constant  ratio  m. 


PKOPOSITION  XII.— THEOKEM. 

32.  An  are  of  a  circle  is  less  than  any  line  which  envelops  it  and  has 
the  same  extremities. 

Let  A  KB  be  an  arc  of  a  circle,  AB  its  chord  . 
and  let  ALB,  AMB,  etc.,  be  any  lines  enveloping 
it  and  terminating  at  A  and  B. 

Of  all  the  lines  AKB,  ALB,  AMB,  etc.  (each 
of  which  includes  the  segment,  or  area,  AKB,  be 
tween  itself  and  the  chord  AB},  there  must  be  at  least  one  minimum, 
or  shortest  line.*  Now,  no  one  of  the  lines  ALB,  AMB,  etc.,  envelop 
ing  AKB,  can  be  such  a  minimum ;  for,  drawing  a  tangent  CKD  to 
the  arc  AKB,  the  line  ACKDB  is  less  than  ACLDB;  therefore 
ALB  is  not  the  minimum  ;  and  in  the  same  way  it  is  shown  that  no 
other  enveloping  line  can  be  the  minimum.  Therefore,  the  arc  AKB 
is  the  minimum. 

33.  Corollary.   The  circumference  of  a  circle  is  less  than  the  perimeter 
of  any  polygon  circumscribed  about  it. 

34.  Scholium.  The  demonstration  is  applicable  when  AKB  is  any 
convex  curve  whatever. 


PKOPOSITION  XIII.— THEOKEM. 

35.  If  the  number  of  sides  of  a  regular  polygon  inscribed  in  a  circle 
be  increased  indefinitely,  the  apothem  of  the  polygon  will  approach  to 
the  radius  of  the  circle  as  its  limit. 

*  If  we  choose  to  admit  the  possibility  of  two  or  more  equal  shortest  lines,  still 
we  say  that  of  all  the  lines,  AKB,  ALB,  etc.,  there  must  be  one  which  is  either 
the  minimum  line,  or  one  of  the  minimum  lines. 


156 


GEOMETRY. 


Let  AB  be  a  side  of  a  regular  polygon  inscribed 
in  the  circle  whose  radius  is  OA ;  and  let  OD  be 
its  apothem. 

In  the  triangle  OAD  we  have  (I.  67), 

OA—  OD  <  AD. 

Now,  by  increasing  the  number  of  sides  of  the  polygon,  the  length 
of  a  side  AB  may  evidently  be  made  as  small  as  we  please,  or  less 
than  any  quantity  that  may  be  assigned.  Hence  AD,  or  %AB,  and 
still  more  OA  —  OD,  which  is  still  less  than  AD,  may  become  less 
than  any  assigned  quantity ;  that  is,  the  apothem  OD  approaches  to 
the  radius  OA  as  its  limit  (28). 


0 


PROPOSITION  XIV.— THEOREM. 

36.  The  circumference  of  a  circle  is  the  limit  to  which  the  perimeters 
of  the  inscribed  and  circumscribed  polygons  approach  when  the  number 
of  their  sides  is  increased  indefinitely;  and  the  area  of  the  circle  is  the 
limit  of  the  areas  of  these  polygons. 

Let  AB  and  CD  be  sides  of  an  inscribed  and  a 
similar  circumscribed  polygon;  let  r  denote  the 
apothem  OE,  R  the  radius  OF,  p  the  perimeter  of 
the  inscribed  polygon,  P  the  perimeter  of  the  cir 
cumscribed  polygon.  Then,  we  have  (10), 

^=  — 

p=--r' 

whence,  by  division  (III.  10), 

P--p___R  —  r  Qrp          =Px(E._r, 
PR  R 

Now  we  have  seen,  in  the  preceding  proposition,  that  by  increasing 
the  number  of  sides  of  the  polygons,  the  difference  R  —  r  may  be 
made  less  than  any  assigned  quantity;  consequently  the  quantity 

£  X  (B r),  or  P  —  p,  may  also  be  made  less  than  any  assigned 

quantity.  But  P  being  always  greater,  and  p  always  less,  than  "the 
circumference  of  the  circle,  the  difference  between  this  circumference 


BOOK    V.  157 

and  either  Porp  is  less  than  the  difference  P  —  p,  and  consequently 
may  also  be  made  less  than  any  assigned  quantity.  Therefore,  the 
circumference  is  the  common  limit  of  P  and  p. 

Again,  let  s  and  S  denote  the  areas  of  two  similar  inscribed  and 
circumscribed  polygons.  The  difference  between  the  triangles  COD 
and  AOB  is  the  trapezoid  CABD,  the  measure  of  which  is 
£(CD  -f  AB)  X  EF\  therefore,  the  difference  between  the  areas  of 
the  polygons  is 


consequently, 

S—  s<PX  (P  —  r). 

Now  by  increasing  the  number  of  sides  of  the  polygons,  the  quantity 
P  X  (P  —  r),  and  consequently  also  S  —  s,  may  be  made  less  than 
any  assigned  quantity.  But  S  being  always  greater,  and  s  always  less, 
than  the  area  of  the  circle,  the  difference  between  the  area  of  the 
circle  and  either  S  or  s  is  less  than  the  difference  S  —  s,  and  conse 
quently  may  also  be  made  less  than  any  assigned  quantity.  There 
fore,  the  area  of  the  circle  is  the  common  limit  of  /S  and  s. 


PBOPOSITION  XV.—  THEOKEM. 

37.   The  circumferences  of  two  circles  are  to  each  other  as  their  radii, 
and  their  areas  are  to  each  other  as  the  squares  of  their  radii. 

Let  R  and  R'  be  the  radii  of 
the  circles,  C  and  C'  their  cir 
cumferences,  S  and  S'  their  areas. 

Inscribe  in  the  two  circles  simi 
lar  regular  polygons  ;  let  P  and 
P'  denote  the  perimeters,  A  and 
A'  the  areas  of  these  polygons; 
then,  the  polygons  being  similar,  we  have  (10), 

P  R  A=-I*L; 

P'~~  R''       'A~~R'2' 

These  relations  remain  the  same  whatever  may  be  the  number  of 
sides  in  the  polygons,  provided  there  is  the  same  number  in  each  (9). 
When  this  number  is  indefinitely  increased,  P  approaches  C  as  its 

14 


158 


GEOMETEY. 


limit,  and  P'  approaches  C'  as  its 
limit  (36) ;  and  since  Pand  P'  are 
in  the  constant  ratio  of  R  to  R', 
their  limits  are  in  the  same  ratio 
(31);  therefore 

C7" 

And  sin'ce  the  limit  of  A  is  S,  and  the  limit  of  A  is  S',  it  follows 
in  the  same  manner  that 


[2] 


38.   Corollary  I.  The  proportion  [1]  is  by  (III.  9)  the  same  as 


C 

C' 


2R' 


and  the  proportion  [2]  is  the  same  as 


v 


0' 


8'       4R'2 

therefore,  the  circumferences  of  circles  are  to  each  other  as  their  diame 
ters,  and  their  areas  are  to  each  other  as  the  squares  of  their  diameters. 

39.  Corollary  II.  Similar  arcs,  as  AB, 
A'B',  are  those  which  subtend  equal  an 
gles  at  the  centres  of  the  circles  to  which 
they   belong;    they   are   therefore   like 
parts  of  their  respective  circumferences, 

and  are  in  the  same  ratio  as  the  circumferences.  Also' the  similar 
sectors  A  OB  and  AO'B'  are  like  parts  of  the  circles  to  which  they 
belong.  Therefore,  similar  arcs  are  to  each  other  as  their  radii,  and 
similar  sectors  are  to  each  other  as  the  squares  of  their  radii. 

40.  Corollary  III.    The  ratio  of  the  circumference  of  a  circle  to  its 
diameter  is  constant;  that  is,  it  is  the  same  for  all  circles.     For,  from 
the  proportion  [3]  we  have 

ri  r^r 

\j  \_/ 

C)  ~D  O  D  / 

Zi±\,         Zi±l 


BOOK    V.  159 

This  constant  ratio  is  usually  denoted  by  TT,  so  that  for  any  circle 
whose  diameter  is  2R  and  circumference  C,  we  have 


41.  Scholium.  The  ratio  it  is  incommensurable  (as  can  be  proved 
by  the  higher  mathematics),  and  can  therefore  be  expressed  in  num 
bers  only  approximately.  The  letter  TT,  however,  is  used  to  symbolize 
its  exact  value. 


PEOPOSITION  XVI.—  THEOKEM. 

42.  The  area  of  a  circle  is  equal  to  half  the  product  of  its  circum 
ference  by  its  radius. 

Let  the  area  of  any  regular  polygon  circum 
scribed  about  the  circle  be  denoted  by  A,  its 
perimeter  by  P,  and  its  apothem  which  is  equal 
to  the  radius  of  the  circle  by  R  ;  then  (22), 


Let  the  number  of  the  sides  of  the  polygon  be  continually  doubled, 
then  A  approaches  the  area  S  of  the  circle  as  its  limit,  and  P  ap 
proaches  the  circumference  C  as  its  limit  ;  but  A  and  P  are  in  the 
constant  ratio  %R  ;  therefore  their  limits  are  in  the  same  ratio  (31), 
and  we  have 

~  =  iJ&«*£=ei0X&  [1J 

C 

43.  Corollary  I.  The  area  of  a  circle  is  equal  to  the  square  of  its 
radius  multiplied  by  the  constant  number  TT.  For,  substituting  for  6 
its  value  2nR  in  [1],  we  have 


44.  Corollary  II.  The  area  of  a  sector  is  equal  to  half  the  product  of 
its  arc  by  the  radius.  For,  denote  the  arc  ab  of  the  sector  a  0  b  by  c, 
and  the  area  of  the  sector  by  s  ;  then,  since  c  and  s  are  like  parts  of 
C  and  8,  we  have  (III.  9), 


160 


=±CX  R;  therefore  s  =  \c  X  R. 

45.  Scholium.  A  circle  may  be  regarded  as  a  regular  polygon  of  an 
infinite  number  of  sides.  In  proving  that  the  circle  is  the  limit  to 
wards  which  the  inscribed  regular  polygon  approaches  when  the 
number  of  its  sides  is  increased  indefinitely,  it  was  tacitly  assumed 
that  the  number  of  sides  is  always  finite.  It  was  shown  that  the  dif 
ference  between  the  polygon  and  the  circle  may  be  made  less  than 
any  assigned  quantity  by  making  the  number  of  sides  sufficiently 
great  ;  but  an  assigned  difference  being  necessarily  a  finite  quantity, 
there  is  also  some  finite  number  of  sides  sufficiently  great  to  satisfy 
the  imposed  condition.  Conversely,  so  long  as  the  number  of  sides 
is  finite,  there  is  some  finite  difference  between  the  polygon  and  the 
circle.  But  if  we  make  the  hypothesis  that  the  number  of  sides  of 
the  inscribed  regular  polygon  is  greater  than  any  finite  number,  that 
is,  infinite,  then  it  must  follow  that  the  difference  between  the  poly 
gon  and  the  circle  is  less  than  any  finite  quantity,  that  is,  zero  ;  and 
consequently,  the  circle  is  identical  with  the  inscribed  polygon  of  an 
infinite  number  of  sides. 

This  conclusion,  it  will  be  observed,  is  little  else  than  an  abridged 
statement  of  the  theory  of  limits  as  applied  to  the  circle  ;  the  abridg 
ment  being  effected  by  the  hypothetical  introduction  of  the  infinite 
into  the  statement. 


PROPOSITION  XVII.— PROBLEM. 

46.  To  compute  the  ratio  of  the  circumference  of  a  circle  to  its  diame 
ter,  approximately. 

FIRST  METHOD,  called  the  METHOD  OF  PERIMETERS.  In  this 
method,  we  take  the  diameter  of  the  circle  as  given  and  compute  the 
perimeters  of  some  inscribed  and  a  similar  circumscribed  regular 
polygon.  We  then  compute  the  perimeters  of  inscribed  and  circum 
scribed  regular  polygons  of  double  the  number  of  sides,  by  Propo 
sition  X.  Taking  the  last-found  perimeters  as  given,  we  compute 
the  perimeters  of  polygons  of  double  the  number  of  sides  by  the  same 
method ;  and  so  on.  As  the  number  of  sides  increases,  the  lengths 


BOOK    V. 


161 


of  the  perimeters  approach  to  that  of  the  circumference  (36)  ;  hence, 
their  successively  computed  values  will  be  successive  nearer  and 
nearer  approximations  to  the  value  of  the  circumference. 

Taking,  then,  the  diameter  of  the  circle  as  given  =  1,  let  us  begin 
by  inscribing  and  circumscribing  a  square.  The  perimeter  of  the 
inscribed  square  =  4  X  i  X  1/2"=  2i/2~(13) ;  that  of  the  circum 
scribed  square  =  4 ;  therefore,  putting 

P=4. 

p  =  2-J/2  ==  2.8284271, 

we  find,  by  Proposition  X.,  for  the  perimeters  of  the  circumscribed 
and  inscribed  regular  octagons, 

=  2p  X  P= 


T*  =  3-°614675- 

Then  taking  these  as  given  quantities,  we  put 

P  =  3.3137085,  p  =  3.0614675, 
and  find  by  the  same  formulae  for  the  polygons  of  16  sides 

P'  =  3.1825979,  p'  =  3.1214452. 

Continuing  this  process,  the  results  will  be  found  as  in  the  following 

TABLE* 


Number 
of  sides. 

Perimeter  of 
circumscribed  polygon. 

Perimeter  of 
inscribed  polygon. 

4 

4.0000000 

2.8284271 

8 

3.3137085 

3.0614675 

16 

3.1825979 

3.1214452 

32 

3.1517249 

3.1365485 

64 

3.1441184 

3.1403312 

128 

3.1422236 

3.1412773 

256 

3.1417504 

3.1415138 

512 

3.1416321 

3.1415729 

1024 

3.1416025 

3.1415877 

2048 

3.1415951 

3.1415914 

4096 

3.1415933 

3.1415923 

8192 

3.1415928 

3.1415926 

*  The  computations  have  been  carried  out  with  ten  decimal  places  in  order  to 
ensure  the  accuracy  of  the  seventh  place  as  given  in  the  table. 
14*  L 


162  GEOMETRY. 

From  the  last  two  numbers  of  this  table,  we  learn  that  the  cir 
cumference  of  the  circle  whose  diameter  is  unity  is  less  than 
3.1415928  and  greater  than  3.1415926 ;  and  since,  when  the  diame 
ter  =  1,  we  have  C  —  TT,  (40),  it  follows  that 

TT  =  3.1415927 

within  a  unit  of  the  seventh  decimal  place. 

SECOND  METHOD,  called  the  METHOD  OP  ISOPERIMETERS.  This 
method  is  based  upon  Proposition  XL  Instead  of  taking  the  diame 
ter  as  given  and  computing  its  circumference,  we  take  the  circum 
ference  as  given  and  compute  the  diameter ;  or  we  take  the  semi- 
circumference  as  given  and  compute  the  radius. 

Suppose  we  assume  the  semi-circumference  -|  C  —  1 ;  then  since 
C  =  2xR,  we  have 

=  ~R==R* 

that  is,  the  value  of  TT  is  the  reciprocal  of  the  value  of  the  radius  of 
the  circle  whose  semi-circumference  is  unity. 

Let  ABCD  be  a  square  whose  semi-perimeter      A          E B 

=  1 ;  then  each  of  its   sides  =  ^-.      Denote  its 
radius  OA  by  R,  and  its  apothem  OE  by  r ;  then 

we  have 

r  =  \       —  0.2500000, 

R  =  £j/2"=  0.3535534. 

Now,  by  Proposition  XL,  we  compute  the  apothem  r'  and  the 
radius  R'  of  the  regular  polygon  of  8  sides  having'  the  same  pe 
rimeter  as  this  square  ;  we  find 

r>  =     -?_±J:  ^0.3017767, 

L 

R'  =  i/R  x  r'  =  0.3266407. 
Again,  taking  these  as  given,  we  put 

r  =  0.3017767,  R  =  0.3266407, 

and  find  by  the  same  formulse,  for  the  apothem  and  radius  of  the 
isoperimetric  regular  polygon  of  16  sides,  the  values 

r'  =  0.3142087,  R'  =  0.3203644. 


BOOK    V.  163 

Continuing  this  process,  the  results  are  found  as  in  the  following 


TABLE. 


Number  of  sides. 

Apothem. 

Radius. 

4 

0.2500000 

0.3535534 

8 

0.3017767 

0.3266407 

16 

0.3142087 

0.3203644 

32 

0.3172866 

0.3188218 

64 

0.3180541 

0.3184376 

128 

0.3182460 

0.3183418 

256 

0.3182939 

0.3183179 

512 

0.3183059 

0.3183119 

1024 

0.3183089 

0.3183104 

2048 

0.3183096 

0.3183100 

4096 

0.3183098 

0.3183099 

8192 

0.3183099 

0.3183099 

Now,  a  circumference  described  with  the  radius  r  is  inscribed  in  the 
polygon,  and  a  circumference  described  with  a  radius  R  is  circum 
scribed  about  the  polygon  ;  and  the  first  circumference  is  less,  while 
the  second  is  greater,  than  the  perimeter  of  the  polygon.  Therefore 
the  circumference  which  is  equal  to  the  perimeter  of  the  polygon  has 
a  radius  greater  than  r  and  less  than  R  •  and  this  is  true  for  each  of 
the  successive  isoperimetric  polygons.  But  the  r  and  R  of  the 
polygon  of  8192  sides  do  not  differ  by  so  much  as  .0000001 ;  there 
fore,  the  radius  of  the  circumference  which  is  equal  to  the  perimeter 
of  the  polygons,  that  is,  to  2,  is  0.3183099  within  less  than  .0000001 ; 
and  we  have 


7T  = 


0.3183099 


=  3.141593 


within  a  unit  of  the  sixth  decimal  place. 

47.  Scholium  I.  Observing  that  in  this  second  method  the  value 
of  r  =  1,  for  the  square,  is  the  arithmetic  mean  of  0  and  £,  and  that 
•^  —  il/2  is  the  geometric  mean  between  \  and  J,  we  arrive  at  the 
following  proposition : 

The  value  of  -  is  the  limit  approached  by  the  successive  numbers  ob 
tained  by  starting  from  the  numbers  0  and  %  and  taking  alternately  the 
arithmetic  mean  and  the  geometric  mean  between  the  two  which  precede. 


164  GEOMETRY. 

48.  Scholium  II.  ARCHIMEDES  (born  287  B.  C.)  was  the  first  to 
assign  an  approximate  value  of  n.  By  a  method  similar  to  the 
above  "  first  method,"  he  proved  that  its  value  is  between  3^  and 
3-ff-,  or,  in  decimals,  between  3.1428  and  3.1408 ;  he  therefore 
assigned  its  value  correctly  within  a  unit  of  the  third  decimal  place. 
The  number  3-f,  or  2T2,  usually  cited  as  Archimedes'  value  of  ?r 
(although  it  is  but  one  of  the  two  limits  assigned  by  him),  is  often 
used  as  a  sufficient  approximation  in  rough  computations. 

METIUS  (A.  D.  1640)  found  the  much  more  accurate  value  f|-f, 
which  correctly  represents  even  the  sixth  decimal  place.  It  is  easily 
remembered  by  observing  that  the  denominator  and  numerator  writ 
ten  consecutively,  thus  113 1 355,  present  the  first  three  odd  numbers 
each  written  twice. 

More  recently,  the  value  has  been  found  to  a  very  great  number 
of  decimals,  by  the  aid  of  series  demonstrated  by  the  Differential 
Calculus.  CLAUSEN  and  DASE  of  Germany  (about  A.  D.  1846),  com 
puting  independently  of  each  other,  carried  out  the  value  to  200 
decimal  places,  and  their  results  agreed  to  the  last  figure.  The 
mutual  verification  thus  obtained  stamps  their  results  as  thus  far 
the  best  established  value  to  the  200th  place.  (See  SCHUMACHER'S 
Asironomisehe  Nachrichten,  No.  589.)  Other  computers  have  car 
ried  the  value  to  over  500  places,  but  it  does  not  appear  that  their 
results  have  been  verified. 

The  value  to  fifteen  decimal  places  is 

TT  =  3.141592653589793. 

For  the  greater  number  of  practical  applications,  the  value  K  =  3.1416 
is  sufficiently  accurate. 


MAXIMA    AND    MINIMA    OF    PLANE    FIGURES. 

49.  Definition.  Among  quantities  of  the  same  kind,  that  which  is 
greatest  is  called  a  maximum ;  that  which  is  least,  a  minimum. 

Thus,  the  diameter  of  a  circle  is  a  maximum  among  all  straight 
lines  joining  two  points  of  the  circumference;  the  perpendicular  is  a 
minimum  among  all  the  lines  drawn  from  a  given  point  to  a  given 
straight  line. 


BOOK    V.  165 

An  enclosed  figure  is  said  to  be  a  maximum  or  a  minimum,  when 
its  area  is  a  maximum  or  a  minimum. 

50.  Definition.  Any  two  figures  are  called  isoperimetric  when 
their  perimeters  are  equal. 


PROPOSITION  XVIIL— THEOREM. 

51.  Of  all  triangles  having  the  same  base  and  equal  areas,  that  which 
is  isosceles  has  the  minimum  perimeter. 

Let  ABC  be  an  isosceles  triangle,  and  ABC 
any  other  triangle  having  the  same  base  and  an 
equal  area;  then,  AB  -f  AC  <  AB  +  AC. 

For,  the  altitudes  of  the  triangles  must  be 
equal  (IV.  15),  and  their  vertices  A  and  A  lie 
in  the  same  straight  line  MN  parallel  to  BC. 
Draw  CND  perpendicular  to  MN,  meeting  BA  produced  in  D ;  join 
AD.  Since  the  angle  NAG  =  ACB  =  ABC=  DAN,  the  right 
triangles  A  CN,  ADN,  are  equal ;  therefore,  AN  is  perpendicular  to 
CD  at  its  middle  point  N,  and  we  have  AD  —AC,  AD  =  AC. 
But  BD  <  A'B  +  AD]  that  is,  AB  +  AC <  AB  +  AC. 

52.  Corollary.    Of  all  triangles  having  the  same  area,  that  which  is 
equilateral  has  the  minimum  perimeter.     For,  the  triangle  having  the 
minimum  perimeter  enclosing  a  given  area  must  be  isosceles  which 
ever  side  is  taken  as  the  base. 


PROPOSITION  XIX.— THEOREM. 

53.  Of  all  triangles  having  the  same  base  and  equal  perimeters,  thai 
which  is  isosceles  is  the  maximum. 

Let  J.jB(7be  an  isosceles  triangle,  and  let  A'BC, 
standing  on  the  same  base  BC,  have  an  equal 
perimeter;  that  is,  let  A'B  +  A'C  =  AB  -f-  AC; 
then,  the  area  of  ABC  is  greater  than  the  area  of 
A'BC. 

For,  the  vertex  A'  must  fall  between  BC  and  the  parallel  MN 
drawn  through  A ;  since,  if  it  fell  upon  MN,  we  should  have,  as  in 
the  preceding  demonstration,  A'B  -f  A'  C  >  AB  -j-  A  C,  and  if  it 


A' 


166  GEOMETRY. 

fell  above  MN,  the  sum  A'B  -f  A'  C  would  be  still  greater.  There 
fore  the  altitude  of  the  triangle  ABC  is  greater  than  that  of  ABC, 
and  hence  also  its  area  is  the  greater. 

54.  Corollary.    Of  all  isoperimetric  triangles,  that  which  is  equilat 
eral  is  the  maximum.     For,  the  maximum  triangle  having  a  given 
perimeter  must  be  isosceles  whichever  side  is  taken  as  the  base. 

PROPOSITION  XX.— THEOREM. 

55.  Of  all  triangles  formed  with  the  same  two  given  sides,  that  in 
which  these  sides  are  perpendicular  to  each  other  is  the  maximum. 

Let  ABC,  ABC,  be  two  triangles  having  the 
sides  AB,  BC,  respectively  equal  to  A'B,  BC; 
then,  if  the  angle  ABC  is  a  right  angle,  the  area 
of  the  triangle  ABC  is  greater  than  that  of  the 
triangle  ABC. 

For,  taking  BC  as  the  common  base,  the  altitude  AB  of  the  tri 
angle  ABC  is  evidently  greater  than  the  altitude  AD  of  the 
triangle  ABC. 


PROPOSITION  XXI.— THEOREM. 

56.    Of  all  isoperimetrie  plane  figures,  the  circle  is  the  maximum. 

1st.  With  a  given  perimeter,  there  may  be  constructed  an  infinite 
number  of  figures  of  different  forms  and  various  areas.  The  area 
may  be  made  as  small  as  we  please  (IV.  35),  but  obviously  cannot 
be  increased  indefinitely.  Therefore,  among  all  the  figures  of  the 
same  perimeter  there  must  be  one  maximum  figure,  or  several  maxi 
mum  figures  of  different  forms  and  equal  areas. 

2d.  Every  closed  figure  of  given  perimeter  containing  a  maximum 
area  must  necessarily  be  convex,  that  is,  such  that  any  straight  line 
joining  two  points  of  the  perimeter  lies  wholly  within  the  figure. 

Let  ACBNA  be  a  non-convex  figure,  the 
straight  line  AB,  joining  two  of  the  points  in  its 
perimeter,  lying  without  the  figure ;  then,  if  the 
re-entrant  portion  ACB  be  revolved  about  the 
line  AB  into  the  position  AC'B,  the  figure 
AC'BNA  has  the  same  perimeter  as  the  first 


C' 


BOOK    V.  167 

figure,  but  a  greater  area.  Therefore,  the  non-convex  figure  cannot 
be  a  maximum  among  figures  of  equal  perimeters. 

3d.  Now  let  ACBFA  be  a  maximum  figure  formed  with  a  given 
perimeter;  then  we  say  that,  taking  any 
point  A  in  its  perimeter  and  drawing  AB 
so  as  to  divide  the  perimeter  into  two  equal 
parts,  this  line  also  divides  the  area  of  the 
figure  into  two  equal  parts.  For,  if  the 
area  of  one  of  the  parts,  as  AFB,  were 
greater  than  that  of  the  other  part,  ACB, 
then,  if  the  part  AFB  were  revolved  upon 

the  line  AB  into  the  position  AF'B,  the  area  of  the  figure  AF'BFA 
would  be  greater  than  that  of  the  figure  A  CBFA,  and  yet  would 
have  the  same  perimeter ;  thus  the  figure  A  CBFA  would  not  be  a 
maximum. 

Hence  also  it  appears  that,  A  CBFA  being  a  maximum  figure, 
AF'BFA  is  also  one  of  the  maximum  figures,  for  it  has  the  same 
perimeter  and  area  as  the  former  figure.  This  latter  figure  is  sym 
metrical  with  respect  to  the  line  AB,  since  by  the  nature  of  the  revo 
lution  about  AB,  every  line  FF'  perpendicular  to  AB,  and  termi 
nated  by  the  perimeter,  is  bisected  by  AB  (1. 140).  Hence  F  and  F' 
being  any  two  symmetrical  points  in  the  perimeter  of  this  figure,  the 
triangles  AFB  and  AF'B  are  equal. 

Now  the  angles  AFB  and  AF'B  must  be  right  angles ;  for  if  they 
were  not  right  angles  the  areas  of  the  triangles  AFB  and  AF'B 
could  be  increased  without  varying  the  lengths  of  the  chords  AF9 
FB,  AF',  F'B  (55),  and  then  (the  segments  AGF,  FEB,  AG'F', 
F'E'B,  still  standing  on  these  chords),  the  whole  figure  would  have 
its  area  increased  without  changing  the  length  of  its  perimeter ;  con 
sequently  the  figure  AF'BFA  would  not  be  a  maximum.  There 
fore,  the  angles  jPand  F'  are  right  angles.  But  Fis  any  point  in  the 
curve  AFB;  therefore,  this  curve  is  a  semi-circumference  (II.  59,  97). 

Hence,  if  a  figure  ACBFA  of  a  given  perimeter  is  a  maximum, 
its  half  AFB,  formed  by  drawing  AB  from  any  arbitrarily  chosen 
point  A  in  the  perimeter,  is  a  semicircle.  Therefore  the  whole  figure 
is  a  circle.* 

*  This  demonstration  is  due  to  STEINER,  Crelle's  Journal  fur  die  reine  und 
angewandte  Mathematik,  vol.  24.  (Berlin,  1842.) 


168 


GEOMETRY. 


x~x 


PROPOSITION  XXII.— THEOREM. 

57.   Of  all  plane  figures  containing  the  same  area,  the  circle  has  the 
minimum  perimeter. 

Let  C  be  a  circle,  and  A  any 
other  figure  having  the  same  area 
as  C',  then,  the  perimeter  of  C  is 
less  than  that  of  A. 

For,  let  B  be  a  circle  having 
the  same  perimeter  as  the  figure 
A ;  then,  by  the  preceding  theo 
rems  A  <  B,  or  C  <  B.  Now,  of 
two  circles,  that  which  has  the  less 
area  has  the  less  perimeter ;  there 
fore,  the  perimeter  of  C  is  less  than  that  of  B,  or  less  than  that  of  A. 


PROPOSITION  XXIII.— THEOREM. 

58.   Of  all  the  polygons  constructed  with  the  same  given  sides,  that  is 
the  maximum  which  can  be  inscribed  in  a  circle. 

Let  P  be  a  polygon  constructed  with  the  sides  a,  b,  c,  d,  e,  and 
inscribed  in  a  circle  S,  and  let  P' 
be  any  other  polygon  constructed 
with  the  same  sides  and  not  inscrip- 
tible  in  a  circle;  then,  P  >  P'. 

For,  upon  the  sides  a,  b,  c,  etc., 
of  the   polygon  P'  construct   cir 
cular  segments  equal  to  those  stand 
ing  on  the  corresponding  sides  of  P.      The  whole  figure   S'  thus 
formed  has  the  same   perimeter  as  the  circle  S;  therefore,  area  of 
S  >  area  of  Sr  (56) ;  subtracting  the  circular  segments  from  both, 
we  have  P  >  P'. 


BOOK     V. 


169 


PROPOSITION  XXIV.— PROBLEM. 

59.  Of  all  isoperimetric  polygons  having  the  same  number  of  sides, 
the  regular  polygon  is  the  maximum. 

1st.  The  maximum  polygon  P,  of  all  the  isope 
rimetric  polygons  of  the  same  number  of  sides  must 
have  its  sides  equal ;  for  if  two  of  its  sides,  as  AB', 
B'C,  were  unequal,  we  could,  by  (53),  substitute  for 
the  triangle  AB' C  the  isosceles  triangle  ABC 
having  the  same  perimeter  as  AB'CsiTid.  a  greater 
area,  and  thus  the  area  of  the  whole  polygon  could  be  increased  with 
out  changing  the  length  of  its  perimeter  or  the  number  of  its  sides. 

2d.  The  maximum  polygon  constructed  with  the  same  number^of 
equal  sides  must,  by  (58),  be  inscriptible  in  a  circle ;  therefore  it 
must  be  a  regular  polygon. 


PROPOSITION  XXV.— THEOREM. 

60.    Of  all  polygons  having  the  same  number  of  sides  and  the  same 
area,  ike  regular  polygon  has  the  minimum  perimeter. 

Let  P  be  a  regular  polygon,  and  M 
any  irregular  polygon  having  the 
same  number  of  sides  and  the  same 
area  as  P;  then,  the  perimeter  of  Pis 
less  than  that  of  M. 

For,  let  N  be  a  regular  polygon 
having  the  same  perimeter  and  the 
same  number  of  sides  as  M;  then,  by 
(59),  M  <  N,  or  P  <  N.  But  of  two 
regular  polygons  having  the  same 
number  of  sides,  that  which  has  the 

less  area  has  the  less  perimeter ;  therefore  the  perimeter  of  P  is  less 
than  that  of  JV,  or  less  than  that  of  M. 
15 


170 


GEOMETKY. 


PEOPOSITION  XXVI.— THEOEEM. 

61.  If  a  regular  polygon  be  constructed  with  a  given  perimeter,  its 
area  will  be  the  greater,  the  greater  the  number  of  its  sides. 

Let  P  be  the  regular  polygon  of  three 
sides,  and  Q  the  regular  polygon  of  four 
sides,  constructed  with  the  same  given 
perimeter.     In  any  side  AB  of  P  take 
any  arbitrary  point  D  •  the  polygon  P 
may  be  regarded  as  an  irregular  poly 
gon  of  four  sides,  in  which  the  sides  AD,  DB,  make  an  angle  with 
each  other  equal  to  two  right  angles  (I.  16) ;   then,  the  irregular 
polygon  P  of  four  sides  is  less  than  the  regular  isoperimetric  polygon 
Q  of  four  sides  (59).     In  the  same  manner  it  follows  that  Q  is  less 
than  the  regular  isoperimetric  polygon  of  five  sides,  and  so  on. 


PEOPOSITION  XXVII.— THEOEEM. 

62.  If  a  regular  polygon  be  constructed  with  a  given  area,  its  perim 
eter  will  be  the  less,  the  greater  the  number  of  its  sides. 

Let  P  and  Q  be  regular  polygons 
having  the  same  area,  and  let  Q  have 
the  greater  number  of  sides ;  then,  the 
perimeter  of  P  will  be  greater  than  that 

of  Q. 

For,  let  R  be  a  regular  polygon  having 
the  same  perimeter  as  Q  and  the  same 
number  of  sides  as  P;   then,  by  (61), 
Q  >  R,  or  P  >  R ;  therefore  the  perimeter  of  P  is  greater  than 
that  of  E,  or  greater  than  that  of  Q. 


GEOMETRY  OF  SPACE. 


BOOK  VI. 

THE  PLANE.    POLYEDRAL   ANGLES. 

1.  DEFINITION.  A  plane  has  already  been  defined  as  a  surface  such 
that  the  straight  line  joining  any  two  points  in  it  lies  wholly  in  the 
surface. 


Thus,  the  surface  MN  is  a  plane,  if,  A  and  B 


being  any  two  points  in  it,  the  straight  line  AB         I  f + — J 

lies  wholly  in  the  surface.  {_ / 

The  plane  is  understood  to  be  indefinite  in 

extent,  so  that,  however  far  the  straight  line  is  produced,  all  its 
points*  lie  in  the  plane.  But  to  represent  a  plane  in  a  diagram,  we 
are  obliged  to  take  a  limited  portion  of  it,  and  we  usually  represent 
it  by  a  parallelogram  supposed  to  lie  in  the  plane. 


DETERMINATION    OF    A    PLANE. 

PKOPOSITION  I.— THEOREM. 
2.   Through  any  given  straight  line  an  infinite  number  of  planes  may 


Let  AB  be  a  given  straight  line.  A 
straight  line  may  be  drawn  in  any  plane, 
and  the  position  of  that  plane  may  be 
changed  until  the  line  drawn  in  it  is 
brought  into  coincidence  with  AB.  We  shall  then  have  one  plane 

171 


172  GEOMETRY. 

passed  through  AB ;  and  this  plane  may  be  turned  upon  AB  as  an 
axis  and  made  to  occupy  an  infinite  number  of  positions. 

3.  Scholium.  Hence,  a  plane  subjected  to  the  single  condition  that 
it  shall  pass  through  a  given  straight  line,  is  not  fixed,  or  deter 
minate,  in  position.     But  it  will  become  determinate  if  it  is  required 
to  pass  through  an  additional  point,  or  line,  as  shown  in  the  next 
proposition. 

A  plane  is  said  to  be  determined  by  given  lines,  or  points,  when  it 
is  the  only  plane  which  contains  such  lines  or  points. 

•PROPOSITION  II.— THEOREM. 

4.  A  plane  is  determined,  1st,  by  a  straight  line  and  a  point  without 
that  line ;  2d,  by  two  intersecting  straight  lines ;  3d,  by  three  points  not 
in  the  same  straight  line ;  4th,  by  two  parallel  straight  lines. 

1st.  A  plane  MN  being  passed  through  a  given  straight  line  AB,  and 
then  turned  upon  this  line  as  an  axis  until  it 
contains  a  given  point.  C  not  in  the  line  AB, 
is  evidently  determined ;  for,  if  it  is  then 
turned  in  either  direction  about  AB,  it  will 
cease  to  contain  the  point  C.  The  plane  is 
therefore  determined  by  the  given  straight 
line  and  the  point  without  it. 

2d.  If  two  intersecting  straight  lines  AB,  AC,  are  given,  a  plane 
passed  through  AB  and  any  point  C  (other  than  the  point  A)  of  AC, 
contains  the  two  straight  lines,  and  is  determined  by  tliese  lines. 

3d.  If  three  points  are  given,  A,  B,  C,  not  in  the  same  straight 
line,  any  two  of  them  may  be  joined  by  a  straight  line,  and  then  the 
plane  passed  through  this  line  and  the  third  point,  contains  the  three 
points,  and  is  thus  determined  by  them. 

4th.  Two  parallel  lines,  AB,  CD,  are  by      A~~ 
definition    (I.   42)    necessarily   in    the   same 
plane,  and  there  is  but  one  plane  containing 
them,  since  a  plane  passed  through  one  of 
them,  AB,  and  any  point  E  of  the  other,  is  determined  in  position. 

5.  Corollary.  The  intersection  of  two  planes  is  a  straight  line. 
For,  the  intersection  cannot  contain  three  points  not  in  the  same 
straight  line,  since  only  one  plane  can  contain  three  such  points.  •- ' 


BOOK      VI.  173 


PEEPENDICULAES    AND   OBLIQUE    LINES    TO    PLANES. 

6.  Definition.  A  straight  line  is  perpendicular  to  a  plane  when  it 
is  perpendicular  to  every  straight  line  drawn  in  the  plane  through 
its  foot,  that  is,  through  the  point  in  which  it  meets  the  plane. 

In  the  same  case,  the  plane  is  said  to  be  perpendicular  to  the  line. 


PEOPOSITION  III.— THEOEEM. 

7.  From  a  given  point  without  a  plane,  one  perpendimlar  to  the  plane 
can  be  drawn,  and  but  one. 

Let  A  be  the  given  point,  and  MN  the  plane. 

If  any  straight  line,  as  AB,  is  drawn  from  A 
to  a  point  B  of  the  plane,  and  the  point  B  is 
then  supposed  to  move  in  the  plane,  the  length 
of  AB  will  vary.  Thus,  if  B  move  along  a 
straight  line  BB'  in  the  plane,  the  distance  AB 
will  vary  according  to  the  distance  of  B  from 
the  foot  C  of  the  perpendicular  AC  let  fall  from  A  upon  BB'. 
Now,  of  all  the  lines  drawn  from  A.  to  points  in  the  plane,  there 
must  be  one  minimum,  or  shortest  line.  There  cannot  be  two  equal 
shortest  lines  ;  for  if  AB  and  AB'  are  two  equal  straight  lines  from 
A  to  the  plane,  each  is  greater  than  the  perpendicular  A  C  let  fall 
from  A  upon  BB' ;  hence  they  are  not  minimum  lines.  There  is 
therefore  one,  and  but  one,  minimum  line  from  A  to  the  plane.  Let 
AP  be  that  minimum  line ;  then,  AP  is  perpendicular  to  any  straight 
line  EF  drawn  in  the  plane  through  its  foot  P.  For,  in  the  plane  of 
the  lines  AP  and  EF,  AP  is  the  shortest  line  that  can  be  drawn 
from  A  to  any  point  in  EF,  since  it  is  the  shortest  line  that  can  be 
drawn  frorn  A  to  any  point  in  the  plane  MN-,  therefore,  AP  is  per 
pendicular  to  EF  (I.  28).  Thus  AP  is  perpendicular  to  any,  that  is. 
to  every,  straight  line  drawn  in  the  plane  through  its  foot,  and  is 
therefore  perpendicular  to  the  plane.  Moreover,  by  the  nature  of 
the  proof  just  given,  AP  is  the  only  perpendicular  that  can  be  drawn 
from  A  to  the  plane  MN. 

8.  Corollary.  At  a  given  point  P  in  a  plane  MNt  a  perpendicular 
can  be  erected  to  the  plane,  and  but  one. 

15*  * 


174 


GEOMETRY. 


A    B 


M> 


For,  let  M'N'  be  any  other  plane,  A'  any  point  without  it,  and 
A'P'  the  perpendicular  from  A'  to 
this  plane.    Suppose  the  plane  M'N' 
to  be  applied  to  the  plane  MN  with 
the  point  P'  upon  P,  and  let  AP  be 
the  position  then  occupied  by  the        I 
perpendicular  A'P'.    We  then  have      *- 
one  perpendicular,  AP,  to  the  plane 

MN,  erected  at  P.  There  can  be  no  other :  for  let  PB  be  any  other 
straight  line  drawn  through  P;  let  the  plane  determined  by  the  two 
lines  PA,  PB,  intersect  the  plane  MN  in  the  line  PC;  then,  since 
APC  is  a  right  angle,  BPC  is  not  a  right  angle,  and  therefore  BP  is 
not  perpendicular  to  the  plane. 

9.  Scholium.  By  the  distance  of  a  point  from  a  plane  is  meant  the 
shortest  distance;  hence  it  is  the  perpendicular  distance  from  the 
point  to  the  plane. 


PROPOSITION  IV.— THEOREM. 

10.  Oblique  lines  drawn  from  a  point  to  a  plane,  at  equal  distances 
from  the  perpendicular,  are  equal;  and  of  two  oblique  lines  unequally 
distant  from  the  perpendicular  the  more  remote  is  the  greater. 

1st.  Let  AB,  AC  be  oblique  lines  from 
the  point  A  to  the  plane  MN,  meeting  the 
plane  at  the  equal  distances  PB,  PC,  from 
the  foot  of  the  perpendicular  AP;  then, 
AB  =  AC.  For,  the  right  triangles  APB, 
APC,  are  equal  (I.  76). 

2d.  Let  AD  meet  the  plane  at  a  dis 
tance,  PD,  from  P,  greater  than  PC;  then, 
AD>AC.  For,  upon  PD  take  PB  = 
PC,  and  join  AB:  then  AD>  AB  (I.  35); 
but  AB  =  A  C;  therefore,  AD>AC. 

11.  Corollary  I.  Conversely,  equal  oblique  lines  from  a  point  to  a 
plane  meet  the  plane  at  equal  distances  from  the  perpendicular;  and 
of  two  unequal  oblique  lines,  the  greater  meets  the  plane  at  the 
greater  distance  from  the  perpendicular. 


BOOK     VI.  175 

12.  Corollary  II.  Equal  straight  lines  from  a  point  to  a  plane  meet 
the  plane  in  the  circumference  of  a  circle  whose  centre  is  the  foot  of 
the  perpendicular  from  the  point  to  the  plane.  Hence  we  derive  a 
method  of  drawing  a  perpendicular  from  a  given  point  A  to  a  given 
plane  MN:  find  any  three  points,  B,  C,  E,  in  the  plane,  equidistant 
from  A,  and  find  the  centre  P  of  the  circle  passing  through  these 
points;  the  straight  line  AP  will  be  the  required  perpendicular. 


PROPOSITION  V.— THEOEEM. 

13.  If  a  straight  line  is  perpendicular  to  each  of  two  straight  lines  at 
their  point  of  intersection,  it  is  perpendicular  to  the  plane  of  those  lines. 

Let  AP  be  perpendicular  to  PB  and  PC, 
at  their  intersection  P;  then,  AP  is  perpen 
dicular  to  the  plane  MN  which  contains  those 
lines. 

For,  let  PD  be  any  other  straight  line 
drawn  through  P  in  the  plane  MN.  Draw 
any  straight  line  BDC  intersecting  PB,  PC 
PD,  in  B,  C,  D-,  produce  AP  to  A'  making 
PA'  =  PA,  and  join  A  and  A'  to  each  of 
the  points  B,  C,  D. 

Since  BP  is  perpendicular  to  A  A',  at  its 
middle  point,  we  have  BA  =  BA ',  and  for  a  like  reason  CA=  CA ' ; 
therefore,  the  triangles  ABC,  A'BC,  are  equal  (I.  80).  If,  then, 
the  triangle  ABC  is  turned  about  its  base  B  C  until  its  plane  coin 
cides  with  that  of  the  triangle  A'BC,  the  vertex  A  will  fall  upon  A' \ 
and  as  the  point  D  remains  fixed,  the  line  AD  will  coincide  with 
A  'D ;  therefore,  D  and  P  are  each  equally  distant  from  the  extremi 
ties  'of  A  A ',  and  DP  is  perpendicular  to  A  A '  or  AP  (I.  41).  Hence 
AP  is  perpendicular  to  any  line  PD,  that  is,  to  every  line,  passing 
through  its  foot  in  the  plane  MN,  and  is  consequently  perpendicular 
to  the  plane. 

14.  Corollary  I.  At  a  given  point  P  of  a  straight  line  AP,  a  plane 
can  be  passed  perpendicular  to  that  line,  and  but  one.    For,  two  per 
pendiculars,  PB,  PC,  being  drawn  to  AP  in  any  two  different  planes 
APB,  APC,  passed  through  AP,  the  plane  of  the  lines  PB,  PC,  will 


176 


GEOMETRY. 


be  perpendicular  to  the  line  AP.  Moreover,  no  other  plane  passed 
through  P  can  be  perpendicular  to  AP;  for,  any  other  plane  not  con 
taining  the  point  C  would  cut  the  oblique  line  AC  in  a  point  C' 
different  from  C,  and  we  should  have  the  angle  APC'  different  from 
APC,  and  therefore  not  a  right  angle. 

15.  Corollary  II.    All  the   perpendiculars 
PB,  PC,  PD,  etc.,  drawn  to  a  line  AP  at  the 
same  point,  lie  in  one  plane  perpendicular  to 
AP.      Hence,  if  an  indefinite  straight  line 
PQ,  perpendicular  to  AP,  be  made  to  revolve, 
always  remaining  perpendicular  to  AP,  it  is 
said  to  generate  the  plane  MN  perpendicular 

to  AP-,  for  the  line  PQ  passes  successively,  during  its  revolution, 
through  every  point  of  this  plane. 

16.  Corollary  III.  Through  any  point  C  without  a  given  straight 
line  AP,  a  plane  can  be  passed  perpendicular  to  AP,  and  but  one. 
For,  in  the  plane  determined  by  the  line  AP  and  the  point  C,  the 
perpendicular  CP  can  be  drawn  to  AP,  and  then  the  plane  generated 
by  the  revolution  of  PC  about  AP  as  an  axis  will,  by  the  preceding 
corollary,  be  perpendicular  to  AP;  and  it  is  evident  that  there  can 
be  but  one  such  perpendicular  plane. 


PROPOSITION  VI.— THEOREM. 

17.  If  from  the  foot  of  a  perpendicular  to  a  plane  a  straight  line  is 
drawn  at  right  angles  to  any  line  of  the  plane,  and  its  intersection  with 
that  line  is  joined  to  any  point  of  the  perpendicular,  Mis  last  line  will 
be  perpendicular  to  the  line  of  the  plane. 

Let  AP  be  perpendicular  to  the  plane 
MN;  from  its  foot  P  let  PD  be  drawn  at 
right  angles  to  any  line  BC  of  the  plane; 
then,  A  being  any  point  in  AP,  the  straight 
line  AD  is  perpendicular  to  BC. 

For,  lay  off  DB  =  DC,  and  join  PB,  PC, 
AB,  AC.  Since  DB  =  DC,  we  have 
PB  =  PC  (I.  30),  and  hence  AB  =  AC 

(10).     Therefore,  A  and  D  being  each  equally  distant  from  B  and 
C,  the  line  AD  is  perpendicular  to  BC  (I.  41). 


BOOK    VI.  177 


PAKALLEL    STRAIGHT    LINES    AND   PLANES. 

18.  Definitions.  A  straight  line  is  parallel  to  a  plane  when  it  can 
not  meet  the  plane  though  both  be  indefinitely  produced. 

In  the  same  case,  the  plane  is  said  to  be  parallel  to  the  line. 
Two  planes  are  parallel  when  they  do  not  meet,  both  being  indefi 
nite  in  extent. 

PROPOSITION  VII.— THEOREM. 

19.  If  two  straight  lines  are  parallel,  every  plane  passed  through  one 
of  them  is  parallel  to  the  other. 

Let  AB  and  CD  be  parallel  lines,  and 
MN  any  plane  passed  through  CD] 
then,  the  line  AB  and  the  plane  MN  &TQ 
parallel. 

For,  the  parallels  AB,  CD,  are  in  the 

same  plane,  A  CDB,  which  intersects  the  plane  MN  in  the  line  CD ; 
and  if  AB  could  meet  the  plane  MN,  it  could  meet  it  only  in  some 
point  of  CD\  but  AB  cannot  meet  CD,  since  it  is  parallel  to  it ;  there 
fore  AB  cannot  meet  the  plane  MN. 

20.  Corollary  I.   Through  any  given  straight  line  HK,  a  plane  can 
be  passed  parallel  to  any  other  given  straight  line  AB. 

For,  in  the  plane  determined  by  AB  and  any 
point  H  of  HK,  let  HL  be  drawn  parallel  to 
AB ;  then,  the  plane  MN,  determined  by  HK 
and  HL,  is  parallel  to  AB. 


21.   Corollary  II.   Through  any  given  point  0,  a  plane  can  be  passed 
parallel  to  any  two  given  straight  lines  AB,  CD,  in  space. 

For,  in  the  plane  determined  by  the 
given  point  0  and  the  line  AB  let  a  Ob  be 
drawn  through  0  parallel  to  AB ;  and  in 
the  plane  determined  by  the  point  0  and 
the  line  CD,  let  cOd  be  drawn  through  0 
parallel  to  CD ;  then,  the  plane  determined 
by  the  lines  ab  and  cd  is  parallel  to  each 

of  the  lines  AB  and  CD. 

M 


178 


GEOMETRY. 


M 


N 


PROPOSITION  VIII.— THEOREM. 

• 

22.  If  a  straight  line  and  a  plane  are  parallel,  the  intersections  of 
the  plane  with  planes  passed  through  the  line  are  parallel  to  that  line 
and  to  each  other. 

Let  the  line  AB  be  parallel  to  the  plane 
MN,  and  let  CD,  EF,  etc.,  be  the  intersec 
tions  of  MN  with  planes  passed  through 
AB ;  then,  these  intersections  are  parallel 
to  AB  and  to  each  other. 

For,  the  line  AB  cannot  meet  CD,  since 
it  cannot  meet  the  plane  in  which  CD  lies ;  and  since  these  lines  are 
in  the  same  plane,  AD,  and  cannot  meet,  they  are  parallel.  For  the 
same  reason,  EF,  OH,  are  parallel  to  AB. 

Moreover,  no  two  of  these  intersections,  as  CD,  EF,  can  meet ;  for 
if  they  met,  their  point  of  meeting  and  the  line  AB  would  be  at 
once  in  two  different  planes,  AD  and  AF,  which  is  impossible  (4). 

23.  Corollary.  If  a  straight  line  AB  is  parallel  to  a  plane  MN,  a 
parallel  CD  to  the  line  AB,  drawn  through  any  point  C  of  the  plane, 
lies  in  the  plane. 

For,  the  plane  passed  through  the  line  AB  and  the  point  C  inter 
sects  the  plane  MN  in  a  parallel  to  AB,  which  must  coincide  with 
CD,  since  there  cannot  be  two  parallels  to  AB  drawn  through  the 
same  point  C. 


PROPOSITION  IX.— THEOREM. 

24.  Planes  perpendicular  to  the  same  straight  line  are  parallel  to 
each  other. 


The  planes  MN,  PQ,  perpendicular  to  the  same 
straight  line  AB,  cannot  meet ;  for,  if  they  met,  we 
should  have  through  a  point  of  their  intersection 
two  planes  perpendicular  to  the  same  straight  line, 
which  is  impossible  (16)  ;  therefore  these  planes  are 
parallel. 


BOOK    VI. 


179 


\ 


PKOPOSITION  X.— THEOKEM. 

25.  The  intersections  of  two  parallel  planes  with  any  third  plane  are 
parallel. 

Let  MN  and  PQ  be  parallel  planes,  and 
AD  any  plane  intersecting  them  in  the 
lines  A B  and  CD ;  then,  AB  and  CD  are 
parallel. 

For,  the  lines  AB  and  CD  cannot  meet, 
since  the  planes  in  which  they  are  situ 
ated  cannot  meet,  and  they  are  lines  in 
the  same  plane  AD]  therefore  they  are 
parallel. 

26.  Corollary.  Parallel  lines  A  C,  BD,  intercepted  between  parallel 
planes  MN,  PQ,  are  equal.     For,  the  plane  of  the  parallels  A  C,  BD, 
intersects  the  parallel  planes  MN,  PQ,  in  the  parallel  lines  AB,  CD-, 
therefore,  the  figure  A  BD  C  is  a  parallelogram,  and  AC=  BD. 


PKOPOSITION  XI.— THEOKEM. 

27.  A  straight  line  perpendicular  to  one  of  two  parallel  planes  is 
perpendicular  to  the  other. 

Let  MN  and  PQ  be  parallel  planes,  and  let  the 
straight  line  AB  be  perpendicular  to  PQ ;  then, 
it  will  also  be  perpendicular  to  MN. 

For,  through  A  draw  any  straight  line  A  C  in 
the  plane  MN,  pass  a  plane  through  AB  and  A  C, 
and  let  BD  be  the  intersection  of  this  plane  with 
PQ.  Then  AC  and  BD  are  parallel  (25);  but 
AB  is  perpendicular  to  BD  (6),  and  consequently 
also  to  AC',  therefore  AB,  being  perpendicular  to  any  line  AC  of 
the  plane  MN,  is  perpendicular  to  the  plane  MN. 

28.  Corollary.   Through  any  given  point  A,  one  plane  can  be  passed  : 
parallel  to  a  given  plane  PQ,  and  but  one.     For,  from  A  a  perpen 
dicular  AB  can  be  drawn  to  the  plane  PQ  (7),  and  then  through  A 

i    n .  r^  >  -  Q 


180  GEOMETRY. 

a  plane  MN  can  be  passed  perpendicular  to  AB  (14)  ;  the  plane  MN 
is  parallel  to  PQ  (24). 

No  other  plane  can  be  passed  through  A  parallel  to  PQ;  for  every 
plane  parallel  to  PQ  must  be  perpendicular  to  the  line  AB  (27),  and 
there  can  be  but  one  plane  perpendicular  to  AB  passed  through  the 
same  point  A  (14). 


PKOPOSITION  XII.— THEOKEM. 

29.  The  locus  of  all  the  straight  lines  drawn  through  a  given  point 
parallel  to  a  given  plane,  is  a  plane  passed  through  the  point  parallel  to 
the  given  plane. 

Let  A   be  the  given  point,  and  PQ  the  given        M 

plane ;  then,  every  straight  line  AB,  drawn  through        /  J^4^   I 

A  parallel  to  the  plane  PQ,  lies  in  the  plane  MN      L :      B] 

passed  through  A  parallel  to  PQ. 

For,  pass  any  plane  through  AB,  intersecting  the 
plane  PQ  in  a  straight  line  CD ;  then  AB  is  paral- 
lei  to  CD  (22).  But  CD  is  parallel  to  the  plane 
MN,  since  it  is  in  the  parallel  plane  PQ  and  can 
not  meet  MN\  therefore,  the  line  AB  drawn  through  the  point  A 
parallel  to  CD  lies  in  the  plane  MN  (23). 

30.  Scholium.  In  the  geometry  of  space,  the  term  locus  has  the 
same  general  signification  as  in  plane  geometry  (I.  40)  ;  only  it  is  not 
limited  to  lines,  but  may,  as  in  this  proposition,  be  extended  to  a 
surface.     In  the  present  case,  the  locus  is  the  assemblage  of  all  the 
points  of  all  the  lines  which  satisfy  the  two  conditions  of  passing 
through  a  given  point  and  being  parallel  to  a  given  plane. 

31.  Corollary.  Since  two  straight  lines  are  sufficient  to  determine 
a  plane  (4),  if  two  intersecting  straight  lines  are  each  parallel  to  a 
given  plane,  the  plane  of  these  lines  is  parallel  to  the  given  plane. 


PKOPOSITION  XIII.— THEOREM. 

32.  If  two  angles,  not  in  the  same  plane,  have  their  sides  respectively 
parallel  and  lying  in  the  same  direction,  they  are  equal  and  their 
planes  are  parallel. 


BOOK    VI.  181 

Let  BA  C,B'A' C',  be  two  angles  lying  in  the 
planes  MN,  M'N'',  and  let  AB,  AC,  be  parallel 

respectively  to  A'B',  AC',  and  in  the  same 

7      7     7   « 

directions.*  /     /    / 

1st.  The  angles  A4.  C  and  JB' J. '  C"  are  equal.         ^/    /     /    / 
For,  through  the  parallels  AB,  A'B',  pass  a        /~  /     1&  I 

\A,I/      \ls, 


plane  AB',  and   through   the   parallels  AC 
A'C',   pass    a    plane  AC',   intersecting    the  #' 

first  in  the  line  A  A'.    Let  BC'  be  any  plane 

parallel  to  AA',  intersecting  the  planes  AB',  AC',  in  the  lines  BB', 
CC',  and  the  planes  MN,  M'N',  in  the  lines  BC,  B'C',  respectively. 
Since  AA'  is  parallel  to  the  plane  BC',  the  intersections  BB',  CC', 
are  parallel  to  A  A'  and  to  each  other  (22)  ;  hence,  the  quadrilaterals 
AB'  and  AC'  are  parallelograms,  and  we  have  AB  —  A'B',  AC= 
A'C',  and  BB'  =  AA'  =  CC'.  Therefore,  BB'  and  CC'  are  equal 
and  parallel,  and  the  quadrilateral  BC'  is  a  parallelogram,  and  we 
have  BC=B'C'.  The  triangles  ABC,  A'B'C',  therefore,  have 
their  three  sides  equal  each  to  each,  and  consequently  the  angles 
BAG  and  B'A'C'  are  equal. 

2d.  The  planes  of  these  angles  are  parallel.  For,  each  of  the 
lines  AB,  AC,  being  parallel  to  a  line  of  the  plane  M'N',  is  parallel 
to  that  plane,  therefore  the  plane  MN  of  these  lines  is  parallel  to  the 
plane  M'N'  (31). 

PKOPOSITION  XIV.— THEOKEM. 

33.  If  one  of  two  parallel  lines  is  perpendicular  to  a  plane,  the  other 
is  also  perpendicular  to  that  plane. 

Let  AB,  A'B',  be  parallel  lines,  and  let 
AB  be  perpendicular  to  the  plane  MN;  then,         ^ 
A'B'  is  also  perpendicular  to  MN.  f 

For,  let  A  and  A'  be  the  intersections  of        / 
these  lines  with  the  plane ;  through  A  draw  ~~~N 

any  line  AC'  in  the  plane  MN,  and  through 
A  draw  A  C  parallel  to  A'C'  and  in  the  same  direction.    The  angles 

*  Two  parallels  AB,  A'B',  lie  in  the  same  direction  when  they  lie  on  the  same 
side  of  the  line  AA'  joining  their  origins  A  and  A'.    Compare  note  (I.  60). 
16 


182 


GEOMETRY. 


BAG,  B'A'C',  are  equal  (32) ;  but  BAG  is  a  right  angle,  since  BA 
is  perpendicular  to  the  plane;  hence,  B'A'C'  is  a  right  angle;  that 
is,  B'A  is  perpendicular  to  any  line  AC'  drawn  through  its  foot  in 
the  plane  MN,  and  is  consequently  perpendicular  to  the  plane. 

34.  Corollary  I.   Two  straight  lines  AB,  A'B',  perpendicular  to  the 
same  plane  MN,  are  parallel  to  each  other.     For,  if  through  any  point 
of  A'B'  a  parallel  to  AB  is  drawn,  it  will  be  perpendicular  to  the 
plane  MN,  since  AB  is  perpendicular  to  that  plane ;  but  through  the 
same  point  there  cannot  be  two  perpendiculars  to  the  plane ;  there 
fore,  the  parallel  drawn  to  AB  coincides  with  A'B'. 

35.  Corollary  II.  If  two  straight  lines  A  and  B 
are  parallel  to  a  third  C,  they  are  parallel  to  each 
other.     For,  let  MN  be  a  plane  perpendicular  to 
C;  then  (33),  A  and  B  are  each  perpendicular  to 
this  plane  and  are  parallel  to  each  other  (34). 


M 


N 


36.  Corollary  III.  Two  parallel  planes  are  everywhere  equally  dis 
tant.  All  perpendiculars  to  one  of  two  parallel  planes  are  also  per 
pendicular  to  the  other  (27) ;  and  since  they  are  parallels  (34)  inter 
cepted  between  parallel  planes,  they  are  equal  (26). 


PROPOSITION  XV.— THEOREM. 

37.  If  two  straight  lines  are  intersected  by  three  parallel  planes,  their 
corresponding  segments  are  proportional. 

Let  AB,  CD,  be  intersected  by  the  parallel 
planes  MN,  PQ,  P/S,  in  the  points  A,  E,  B,  and 
C,F,D-,  then, 

AE       CF 
EB  ~  FD 

For,  draw  AD  cutting  the  plane  PQ  in  G,  and 
join  EG  and  FG.     The  plane  of  the  lines  AB, 
AD,  cuts  the  parallel  planes  PQ  and  P/S  in  the 
lines  EG  and  BD ;  therefore,  EG  and  BD  are  parallel  (25,),  and  we 
have  (III.  15), 

AE      AG 

EB  ~~~  GD 


BOOK    VI.  183 

The  plane  of  the  lines  DA  and  D  C  cuts  the  parallel  planes  MN  and 
PQ  in  the  lines  A  C  and  GF-,  therefore,  A  C  and  GF  are  parallel, 
and  we  have 

AG  _  CF 

GD  ~  FD 

Comparing  these  two  proportions,  we  obtain 

~EB  =  JS 

DIEDKAL  ANGLES.— ANGLE  OF  A  LINE  AND  PLANE,  ETC. 

38.  Definition.  When  two  planes  meet  and  are  terminated  by  their 
common  intersection, \  they  form  a  diedral  angle. 

Thus,  the  planeTAE,  AF,  meeting  in  AB,  and  ter 
minated  by  AB,  form  a  diedral  angle. 

The  planes  AE,  AF,  are  called  the  faces,  and  the 
line  AB  the  edge,  of  the  diedral  angle. 

A  diedral  angle  may  be  named  by  four  letters,  one 
in  each  face  and  two  on  its  edge,  the  two  on  the  edge  being  written 
between  the  other  two ;  thus,  the  angle  in  the  figure  may  be  named 
DABC. 

When  there  is  but  one  diedral  angle  formed  at  the  same  edge,  it 
may  be  named  by  two  letters  on  its  edge ;  thus,  in  the  preceding 
figure,  the  diedral  angle  DABC  may  be  named  the  diedral  angle 
AB. 

39.  Definition.  The  angle  CAD  formed  by  two  straight  lines  AC, 
AD,  drawn,  one  in  each  face  of  the  diedral  angle,  perpendicular  to 
its  edge  AB  at  the  same  point,  is  called  the  plane  angle  of  the  diedral 
angle. 

The  plane  angle  thus  formed  is  the  same  at  whatever  point  of  the 
edge  of  the  diedral  angle  it  is  constructed.  Thus,  if  at  B,  we  draw 
BE  and  BFin  the  two  faces  respectively,  and  perpendicular  to  AB, 
the  angle  EBF  is  equal  to  the  angle  CAD,  since  the  sides  of  these 
angles  are  parallel  each  to  each  (32). 

It  is  to  be  observed  that  the  plane  of  the  plane  angle  CAD  is 
perpendicular  to  the  edge  AB  (13)  ;  and  conversely,  a  plane  perpen 
dicular  to  the  edge  of  a  diedral  angle  cuts  its  faces  in  lines  which 


184 


GEOMETRY. 


C" 


Et 


are  perpendicular  to  the  edge  and  therefore  form  the  plane  angle  of 
the  diedral  angle. 

40.  A  diedral  angle  DABC  may  be  conceived  to  be  generated  by 
a  plane,  at  first  coincident  with  a  fixed  plane  AE,  revolving  upon 
the  line  AB  as  an  axis  until  it  comes  into  the  position  AF.     In  this 
revolution,  a  straight  line  CA,  perpendicular  to  AB,  generates  the 
plane  angle  CAD. 

41.  Definition.  Two  diedral  angles  are  equal  when  they  can  be 
placed  so  that  their  faces  shall  coincide. 

Thus,  the  diedral  angles  CABDt 
C'A'B'D',  are  equal,  if,  when  the  edge 
A'B'  is  applied  to  the  edge  AB  and 
the  face  A'F'  to  the  face  AF,  the  face 
A'E'  also  coincides  with  the  face  AE. 

Since  the  faces  continue  to  coincide 
when  produced  indefinitely,  it  is  apparent  that  the  magnitude  of  the 
diedral  angle  does  not  depend  upon  the  extent  of  its  faces,  but  only 
upon  their  relative  position. 

Two  diedral  angles  are  evidently  equal  when  their  plane  angles 
are  equal. 

42.  Definition.  Two  diedral  angles  CABD,  DABE, 
which  have  a  common  edge  AB  and  a  common  plane 
BD  between  them,  are  called  adjacent. 

Two  diedral  angles  are  added  together  by  placing 
them  adjacent  to  each  other.  Thus,  the  diedral  angle 
CABE  is  the  sum  of  the  two  diedral  angles  CABD 
and  DABE. 

43.  Definition.   When  a  plane    CAB  meets 
another  MN,   forming  two  equal  adjacent  die 
dral  angles,  CABMand  CABN,  each  of  these 
angles  is  called  a  right  diedral  angle,  and  the 
plane  CAB  is  perpendicular  to  the  plane  MN. 

It  is  evident  that  in  this  case  the  plane 
angles  CDE,  CDF,  of  the  two  equal  diedral 
angles,  are  right  angles. 

Through  any  straight  line  AB  in  a  plane  MN,  a  plane  CAB  can 
be  passed  perpendicular  to  the  plane  MN.  The  proof  is  similar  to 
that  of  the  corresponding  proposition  in  plane  geometry  (I.  9). 


7 


7 


N 


BOOK      VI. 


185 


PROPOSITION  XVI.— THEOREM. 

44.  Two  diedral  angles  are  in  the  same  ratio  as  their  plane  angles. 
Let  CARD  and  GEFHbe  two  die 
dral  angles ;  and  let  CAD  and  GEH 

be  their  plane  angles. 

Suppose  the  plane  angles  have  a 
common  measure,  contained,  for  exam 
ple,  5  times  in  CAD  and  3  times  in 
GEH;  the  ratio  of  these  angles  is 
then  5  :  3.  Let  straight  lines  be  drawn 

from  the  vertices  of  these  angles,  dividing  the  angle  DA  C  into  5 
equal  parts,  and  the  angle  HEG  into  3  equal  parts,  each  equal  to 
the  common  measure ;  let  planes  be  passed  through  the  edge  AB  and 
the  several  lines  of  division  of  the  plane  angle  CAD,  and  also  planes 
through  the  edge  EF  and  the  several  lines  of  division  of  the  plane 
angle  GEH.  The  given  diedral  angles  are  thus  divided  into  partial 
diedral  angles  which  are  all  equal  to  each  other  since  their  plane 
angles  are  equal.  The  diedral  angle  CABD  contains  5  of  these 
partial  angles,  and  the  diedral  angle  GEFH  contains  3  of  them ; 
therefore,  the  given  diedral  angles  are  also  in  the  ratio  5:3;  that  is, 
they  are  in  the  same  ratio  as  their  plane  angles. 

The  proof  is  extended  to  the  case  in  which  the  given  plane  angles 
are  incommensurable,  by  the  method  exemplified  in  (II.  51). 

45.  Corollary  I.  Since  the  diedral  angle  is  proportional  to  its  plane 
angle  (that  is,  varies  proportionally  with  it),  the  plane  angle  is  taken 
as  the  measure  of  the  diedral  angle,  just  as  an  arc  is  taken  as  the  mea 
sure  of  a  plane  angle.     Thus,  a  diedral  angle  will  be  expressed  by 
45°  if  its  plane  angle  is  expressed  by  45°,  etc. 

46.  Corollary  II.   The  sum  of  two  adjacent  die 
dral  angles,  formed  by  one  plane  meeting  another, 
is  equal  to  two  right  diedral  angles.    For,  the  sum 
of  the  plane  angles  which  measure  them  is  equal 
to  two  right  angles. 

In  a  similar  manner,  a  number  of  properties  of 
diedral  angles  can  be  proved,  which  are  analo 
gous  to  propositions  relating  to  plane  angles. 
The  student  can  establish  the  following : 

16* 


186 


GEOMETRY. 


Opposite  or  vertical  diedral  angles  are  equal;   as    CAJBN  and 
DABM,  in  the  preceding  figure. 


When  a  plane  intersects  two  parallel  planes, 
the  alternate  diedral  angles  are  equal,  and  the 
corresponding  diedral  angles  are  equal;  (the 
terms  alternate  and  corresponding  having  sig 
nifications  similar  to  those  given  in  plane 
geometry.) 


Two  diedral  angles  which^  have  jheir  faces  respectively  parallel,  or 
respectively  perpendicular  to  each  other,  are  either  equal  or  supple 
mentary. 


I/I 


PEOPOSITION  XVII.— THEOEEM. 

47.  If  a  straight  line  is  perpendicular  to  a  plane,  every  plane  passed 
through  the  line  is  also  perpendicular  to  that  plane. 

Let  AB  be  perpendicular  to  the  plane  MN; 
then,  any  plane  PQ,  passed  through  AB,  is  also 
perpendicular  to  MN. 

For,  at  B  draw  BC,  in  the  plane  MN,  perpen-        , 
dicular  to  the  intersection  BQ.     Since  AB  is  per-       / 
pendicular  to  the  plane  MN,  it  is  perpendicular  to     ,jf 
BQ  and  BC;  therefore,  the  angle  ABC  is  the 
plane  angle  of  the  diedral  angle  formed  by  the  planes  PQ  and  MN; 
and  since  the  angle  AB  C  is  a  right  angle,  the  planes  are  perpendicu 
lar  to  each  other. 


N 


48.  Corollary.  If  A 0,  BO  and  CO,  are 
three  straight  lines  perpendicular  to  each 
other  at  a  common  point  0,  each  is  per 
pendicular  to  the  plane  of  the  other  two, 
and  the  three  planes  are  perpendicular  to 
each  other. 


*4 


BOOK    VI.  187 


PEOPOSITION  XVIII.— THEOEEM. 

49.  If  two  planes  are  perpendicular  to  each  other,  a  straight  line 
drawn  in  one  of  them,  perpendicular  to  their  intersection,  is  perpendicu 
lar  to  the  other. 

Let  the  planes  PQ,  and  MN  be  perpendicular  to  each  other ;  and 
at  any  point  B  of  their  intersection  B  Q,  let  BA 
be  drawn,  in  the  plane  PQ,  perpendicular  to  BQ; 
then,  BA  is  perpendicular  to  the  plane  MN. 

For,  drawing  B  C,  in  the  plane  MN,  perpendicu 
lar  to  BQ,  the  angle  ABC  is  a  right  angle,  since  it 
is  the  plane  angle  of  the  right  diedral  angle  formed 
by  the  two  planes ;  therefore,  AB,  perpendicular  to 
•the  two  straight  lines  BQ,  BC,  is  perpendicular  to  their  plane 


50.  Corollary  I.  If  two  planes,  PQ  and  MN,  are  perpendicular  to 
each  other,  a  straight  line  BA  drawn  through  any  point  B  of  their 
intersection  perpendicular  to  one  of  the  planes  MN9  will  lie  in  the 
other  plane  PQ  (8). 

51.  Corollary  II.  If  two  planes,  PQ  and  MN,  are  perpendicular 
to  each  other,  a  straight  line  drawn  from  any  point  A  of  PQ,  per 
pendicular  to  MN,  lies  in  the  plane  PQ  (7). 

PKOPOSITION  XIX.— THEOEEM. 

52.  Through  any  given  straight  line,  a  plane  can  be  passed  perpen 
dicular  to  any  given  plane. 

Let  AB  be  the  given  straight  line  and  MN  the 
given  plane.     From  any  point  A  of  AB  let  A  C 
be  drawn  perpendicular  to  MN,  and  through  AB        3r— 
and  A  C  pass  a  plane  AD.     This  plane  is  perpen-       /  c 
dicular  to  MN  (47).  * ^ 

Moreover,  since,  by  (51),  any  plane  passed 
through  AB  perpendicular  to  MN  must  contain  the  perpendicular 
A  C,  the  plane  AD  is  the  only  plane  perpendicular  to  MN  that  can 
be  passed  through  AB,  unless  AB  is  itself  perpendicular  to  MN,  in 
which  case  an  infinite  number  of  planes  can  be  passed  through  it 
perpendicular  to  MN  (47). 


188 


GEOMETRY. 


PEOPOSITION  XX.— THEOEEM. 

53.  If  two  intersecting  planes  are  each  perpendicular  to  a  third  plane, 
their  intersection  is  also  perpendicular  to  that  plane. 

Let  the  planes  PQ,  JR/S,  intersecting  in  the 
line  AB,  be  perpendicular  to  the  plane  MN; 
then,  AB  is  perpendicular  to  the  plane  MN. 

For,  if  from  any  point  A  of  AB,  a  perpen 
dicular  be  drawn  to  MN,  this  perpendicular         I 
will  lie  in  each  of  the  planes  PQ  and  RS  (50),        /_ 
and  must  therefore  be  their  intersection  AB. 

54.  Scholium.  This  proposition  may  be  otherwise  stated  as  follows : 
If  a  plane  (MN)  is  perpendicular  to  each  of  two  intersecting  planes 
(PQ  and  RS\  it  is  perpendicular  to  the  intersection  (AB)  of  those 
planes. 


1 


PEOPOSITION  XXI.— THEOEEM. 

55.  Every  point  in  the  plane  which  bisects  a  diedral  angle  is  equally 
distant  from  the  faces  of  that  angle. 

Let  the  plane  AM  bisect  the 
diedral  angle  CABD\  let  P  be 
any  point  in  this  plane ;  PE  and 
PF  the  perpendiculars  from  P 
upon  the  planes  AB  G  and  ABD ; 
then,  PE  =  PF. 

For,  through  PE  and  PF  pass 
a  plane,  intersecting  the  planes 
ABC  and  ABD  in  OE  and  OF', 
join  PO.  The  plane  PEF  is  per 
pendicular  to  each  of  the  planes  ABC,  ABD  (47),  and  consequently 
perpendicular  to  their  intersection  AB  (54).  Therefore  the  angles 
POE  and  POF  measure  the  diedral  angles  MABC  and  MABD, 
which  by  hypothesis  are  equal.  Hence  the  right  triangles  POE  and 
POF  are  equal  (I.  83),  and  PE  =  PF. 


BOOK    VI. 


189 


56.  Definitions.  The  projection  of  a  point  A 
upon  a  plane  MN,  is  the  foot  a  of  the  perpen 
dicular  let  fall  from  A  upon  the  plane. 

The  projection  of  a  line  ABCDE .  . . ,  upon  a 
plane  MN,  is  the  line  abcde . . .  formed  by  the 
projections  of  all  the  points  of  the  line 
ABCDE. . .  upon  the  plane. 


PKOPOSITION  XXII.— THEOKEM. 

57.  The  projection  of  a  straight  line  upon  a  plane  is  a  straight  line. 
Let  AB  be  the  given  straight  line,  and 

MN  the  given  plane.  The  plane  Ab,  passed 
through  AB  perpendicular  to  the  plane  MN, 
contains  all  the  perpendiculars  let  fall  from 
points  of  AB  upon  MN  (50) ;  therefore,  these 
perpendiculars  all  meet  the  plane  MN  in  the 
intersection  ab  of  the  perpendicular  plane 
with  MN.  The  projection  of  AB  upon  the 
plane  'MN  is,  consequently,  the  straight  line  ab. 

58.  Scholium.  The  plane  Ab  is  called  the  projecting  plane  of  the 
straight  line  AB  upon  the  plane  MN. 


PKOPOSITION  XXIII.— THEOKEM. 

59.  The  acute  angle  which  a  straight  line  makes  with  its  own  pro 
jection  upon  a  plane,  is  the  least  angle  which  it  makes  ivith  any  line  of 
that  plane.. 

Let  Ba  be  the  projection  of  the  straight  line 
BA  upon  the  plane  MN,  the  point  B  being 
the  point  of  intersection  of  the  line  BA  with 
the  plane;  let  BC  be  any  other  straight  line 
drawn  through  B  in  the  plane ;  then,  the  angle 
ABa  is  less  than  the  angle  ABC. 

For,  take  BC=  Ba,  and  join  AC.  In  the 
triangles  ABa,  ABC,  we  have  AB  common,  and  Ba  =  BC,  but 
Aa  <C  AC,  since  the  perpendicular  is  less  than  any  oblique  line; 
therefore,  the  angle  ABa  is  less  than  the  angle  ABC  (I.  85). 


190 


G  E  O  M  E  T  K  Y. 


/   i    i  X    i    / 
/    ;   X°      d  I 


60.  Definition.  The  acute  angle  which  a  straight  line  makes  with 
its  own  projection  upon  a  plane  is  called  the  inclination  of  the  line  to 
the  plane,  or  the  angle  of  the  line  and  plane. 

61.  Definition.  Two  straight  lines  AB, 
CD,  not  in  the  same  plane,  are  regarded 
as  making  an  angle  with  each  other  which 
is  equal  to  the  angle  between  two  straight 
lines  Ob,  Od,  drawn  through  any  point  0 
in  space,  parallel  respectively  to  the  two 
lines  and  in  the  same  directions. 

JM 

Since  every  straight  line  has  two  oppo 
site  directions  (I.  4),  the  angle  which  one  line  makes  with  another 
is  either  acute  or  obtuse,  according  to  the  directions  considered. 
Thus,  if  Ob  is  drawn  in  the  direction  expressed  by  AB  (that  is,  on 
the  same  side  of  a  straight  line  joining  A  and  0~),  and  if  Od  is  drawn 
in  the  direction  expressed  by  CD,  then  dOb  is  equal  to  the  angle 
which  CD  makes  with  AB ;  but  if  Oa  is  drawn  in  the  direction  ex 
pressed  by  BA  (which  is  the  opposite  of  AB},  while  Od  is  still  in 
the  direction  of  CD,  then  d  Oa  is  equal  to  the  angle  which  CD  makes 
with  BA. 

If  MN  is  any  plane  parallel  to  the  two  lines  AB,  CD  (21),  then 
the  angle  of  these  lines  is  the  same  as  the  angle  of  their  projections 
ab,  cd,  upon  this  plane. 

62.  From  the  preceding  definition,  it  follows  that  when  a  straight 
line  is  perpendicular  to  a  plane,  it  is  perpendicular  to  all  the  lines  of  the 
plane,  whether  the  lines  pass  through  its  foot  or 

not.  For,  let  AB  be  perpendicular  to  the  plane 
MN9  and  CD  any  line  of  the  plane.  At  any 
point  B'  in  CD,  let  A'B'  be  drawn  perpendicu 
lar  to  the  plane;  then,  A'B'  being  parallel  to 
AB,  the  right  angle  A' B'  C  is  equal  to  the  angle 
of  the  lines  AB  and  CD,  that  is,  AB  is  perpen 
dicular  to  CD. 


B' 


BOOK    VI. 

PEOPOSITION  XXIV.— THEOEEM. 

63.  Two  straight  lines  not  in  the  same  plane  being  given :  1st,  a 
common  perpendicular  to  the  two  lines  can  be  drawn ;  2d,  but  one  such 
common  perpendicular  can  be  drawn ;  3d,  the  common  perpendicular  is 
the  shortest  distance  between  the  two  lines. 

Let  AB  and  CD  be  the  given  straight 
lines. 

1st.  Through  one  of  the  given  lines, 
say  AB,  pass  a  plane  MN,  parallel  to 
the  other  (20)  ;  let  cd  be  the  projection 
of  CD  upon  this  plane.  Then,  cd  will 
be  parallel  to  CD  (22),  and  therefore 
not  parallel  to  AB ;  hence  it  will  meet 

AB  in  some  point  c.  At  c  draw  cC  perpendicular  to  cd  in  the  pro 
jecting  plane  Cd ;  then  Cc  is  a  common  perpendicular  to  AB  and 
CD. 

For,  CD  and  cd  being  parallel,  Cc  drawn  perpendicular  to  cd  is 
perpendicular  to  CD.  Also,  since  Cc  is  the  line  which  projects  the 
point  C  upon  the  plane  MN,  it  is  perpendicular  to  that  plane,  and 
therefore  perpendicular  to  AB.  » 

2d.  The  line  Cc  is  the  only  common  perpendicular.  For,  if  an 
other  line  EF,  drawn  between  AB  and  CD,  could  be  perpendicular 
to  AB  and  CD,  it  would  be  perpendicular  also  to  a  line  FG  drawn 
parallel  to  CD  in  the  plane  MN,  and  consequently  perpendicular  to 
the  plane  MN;  but  EH,  drawn  in  the  plane  Cd,  parallel  to  Cc,  is 
perpendicular  to  the  plane  MN;  hence  we  should  have  two  perpen 
diculars  from  the  point  E  to  the  plane  MN,  which  is  impossible. 

3d.  The  common  perpendicular  Cc  is  the  shortest  distance  between 
AB  and  CD.  For,  any  other  distance  EF  is  greater  than  the  per 
pendicular  EH,  or  than  its  equal  Cc. 

64.  Scholium.  The  preceding  construction  furnishes  also  the  angle 
between  AB  and  CD,  namely,  the  angle  Bed. 


192 


GEOMETRY. 


POLYEDRAL    ANGLES. 

65.  Definition.  When  three  or  more  planes  meet  in  a  common 
point,  they  form  a  polyedral  angle. 

Thus,  the  figure  S-ABCD,  formed  by  the 
planes  ASB,  BSC,  CSD,  DSA,  meeting  in  the 
common  point  S,  is  a  polyedral  angle. 

The  point  S  is  the  vertex  of  the  angle ;  the 
intersections  of  the  planes,  SA,  SB,  etc.,  are  its 
edges;  the  portions  of  the  planes  bounded  by 
the  edges  are  its  faces;  the  angles  ASB,  BSC,  etc.,  formed  by  the 
edges,  are  its  face  angles. 

A  triedral  angle  is  a  polyedral  angle  having  but  three  faces,  which 
is  the  least  number  of  faces  that  can  form  a  polyedral  angle. 

66.  Definition.  Two  polyedral  angles  are  equal  when  they  can  be 
applied  to  each  other  so  as  to  coincide  in  all  their  parts. 

Since  two  equal  polyedral  angles  coincide  however  far  their  edges 
and  faces  are  produced,  the  magnitude  of  a  polyedral  angle  does  not 
depend  upon  the  extent  of  its  faces ;  but  in  order  to  represent  the 
angle  clearly  in  a  diagram  we  usually  pass  a  plane,  as  ABCD,  cut 
ting  all  its  faces  in  -straight  lines  AB,  BC,  etc. ;  and  by  the  face  ASB 
is  not  meant  the  triangle  ASB,  but  the  indefinite  surface  included 
between  the  lines  SA  and  SB  indefinitely  produced. 

67.  Definition.  A  polyedral  angle  S-ABCD  is  convex,  when  any 
section,  ABCD,  made  by  a  plane  cutting  all  its  faces,  is  a  convex 
polygon  (I.  95). 

68.  Symmetrical  polyedral  angles. 
BS,  etc.,  through  the  vertex  S, 

we  obtain  another  polyedral 
angle  S-A'B'C'D',  which  is 
symmetrical  with  the  first,  the 
vertex  S  being  the  centre  of 
symmetry. 

If  we  pass  a  plane  A'B'C'D', 
parallel  to  ABCD,  so  as  to 
make  SA'  =  SA,  we  shall  also 
have  SB'  =  SB,  SCr  =  SC, 
etc. ;  for  we  may  suppose  a  third 
parallel  plane  passing  through 


If  we  prodlice  the  edges  AS, 


BOOK     VI.  193 

S,  and  then  A  A',  BBf,  etc.,  being  divided  proportionally  by  three 
parallel  planes  (37),  if  any  one  of  them  is  bisected  at  S,  the  others 
are  also  bisected  at  that  point.  The  points  A',  B',  etc.,  are,  then, 
symmetrical  with  A,  B,  etc.,  the  definition  of  symmetry  in  a  plane 
(I.  138),  being  extended  to  symmetry  in  space. 

The  two  symmetrical  polyedral  angles  are  equal  in  all  their  parts ; 
for  their  face  angles,  ASB  and  A' SB',  BSC  and  B'SC',  are  equal, 
each  to  each,  being  vertical  plane  angles ;  and  the  diedral  angles  at 
the  edges  SA  and  SA',  SB  and  SB',  etc.,  are  equal,  being  vertical 
diedral  angles  formed  by  the  same  planes.  But  the  equal  parts  are 
arranged  in  inverse  order  in  the  two  figures,  as  will  appear  more 
plainly,  if  we  turn  the  polyedral  angle  S-A'B'C'D'  about,  until 
the  polygon  A'B'C'D'  is  brought  into  the  same  plane  with  ABCD, 
the  vertex  /S  remaining  fixed ;  the  polygon  A'B'C'D'  is  then  in  the 
position  abed,  and  it  is  apparent  that  while  in  the  polyedral  angle 
S-ABCD  the  parts  ASB,  BSC,  etc.,  succeed  each  other  in  the 
order  from  right  to  left,  their  corresponding  equal  parts  aSb,  bSc,  etc., 
in  the  polyedral  angle  S-abcd  succeed  each  other  in  the  order  from 
left  to  right.  The  two  figures,  therefore,  cannot  be  made  to  coincide 
by  superposition,  and  are  not  regarded  as  equal  in  the  strict  sense 
of  the  definition  (I.  75),  but  are  said  to  be  equal" by  symmetry. 


PEOPOSITION  XXV.— TIIEOBEM. 

69.   The  sum  of  any  two  face  angles  of  a  triedral  angle  is  greater 
than  the  third. 

The  theorem  requires  proof  only  when  the  third  angle  considered 
is  greater  than  each  of  the  others. 

iietS-ABC  be  a  triedral  angle  in  which  the 
face  angle  ASC  is  greater  than  either  ASB  or 
BSC-,  then,  ASB  +  BSC>  ASC. 

For,  in  the  face  ASC  draw  SD  making  the 
angle  ASD  equal  to  ASB,  and  through  any  point 
D  of  SD  draw  any  straight  line  ADC  cutting  SA 
and  SO;  take  SB  =  SD,  and  join  AB,  BC. 

The  triangles  ASD  and  ASB  are  equal,  by  the  construction  (I.  76), 
whence  AD  ~  AB.     Now,  in  the  triangle  ABC,  we  have 
17  N. 


194  GEOMETRY. 

AS  +  BC>  AC, 

and  subtracting  the  equals  AB  and  AD, 

BODC-, 

therefore,  in  the  triangles  BSC  and  DSC,  we  have  the  angle  BSC  > 
DSC  (I.  85),  and  adding  the  equal  angles  ASB  and  ASD,  we  have 
ASB  +  BSOASC. 


PROPOSITION  XXVI.—  THEOKEM. 

70.   The  sum  of  the  face  angles  of  any  convex  polyedral  angle  is  less 
than  four  right  angles. 

Let  the  polyedral  angle  S  be  cut  by  a  plane, 
making  the  section  ABCDE,  by  hypothesis,  a 
convex  polygon.  From  any  point  0  within  this 
polygon  draw  OA,  OB,  OC,  OD,  OE. 

The  sum  of  the  angles  of  the  triangles  ASB, 
B8C,  etc.,  which  have  the  common  vertex  8,  is 
equal  to  the  sum  of  the  angles  of  the  same  num 
ber  of  triangles  AOB,  BOG,  etc.,  which  have  the  common  vertex 
0.  But  in  the  triedral  angles  formed  at  A,  B,  C,  etc.,  by  the  faces 
of  the  polyedral  angle  and  the  plane  of  the  polygon,  we  have  (69). 

SAE  +  SAB  >  EAB, 
SBA  +  SBC  >  ABC,  etc.; 

hence,  taking  the  sum  of  all  these  inequalities,  it  follows  that  the 
sum  of  the  angles  at  the  bases  of  the  triangles  whose  vertex  is  S  is 
greater  than  the  sum  of  the  angles  at  the  bases  of  the  triangles 
whose  vertex  is  0  ;  therefore,  the  sum  of  the  angles  at  S  is  less  than 
the  sum  of  the  angles  at  0,  that  is,  less  than  four  right  angles. 

PKOPOSITION  XXVII.—  THEOREM. 

71.  Two  triedral  angles  are  either  equal  or  symmetrical,  when  the 
three  face  angles  of  one  are  respectively  equal  to  the  three  face  angles  of 
the  other. 

In  the  triedral  angles  S  and  s,  let  ASB  =  asb,  ASC  =  asc,  and 


BOOK     VI. 


195 


BSC  =  bsc;   then,  the  diedral  angle  SA  is  equal  to  the  diedral 
angle  sa. 


On  the  edges  of  these  angles  take  the  six  equal  distances  SA,  SB, 
SC,  sa,  sb,  sc,  and  draw  AB,  BC,  AC,  ab,  be,  ae.  The  isosceles  tri 
angles  SAB  and  sab  are  equal,  having  an  equal  angle  included  by 
equal  sides,  hence  AB  =  ab',  and  for  the  same  reason,  BC  =  be, 
AC=  ac;  therefore,  the  triangles  ABC  and  abe  are  equal. 

At  any  point  D  in  SA,  draw  DE  in  the  face  ASB  and  DF  in  the 
face  ASC,  perpendicular  to  S. A  ;  these  lines  meet  AB  and  AC, 
respectively,  for,  the  triangles  ASB  and  ASC  being  isosceles,  the 
angles  SAB  and  SAC  are  acute ;  let  E and  F be  the  points  of  meet 
ing,  and  join  EF.  Now  on  sa  take  sd  =  SD,  and  repeat  the  same 
construction  in  the  triedral  angle  s. 

*f  he  triangles  ADE  and  ade  are  equal,  since  AD  =  ad,  and  the 
angles  at  A  and  D  are  equal  to  the  angles  at  a  and  d;  hence, 
AE  =  ae  and  DE  =  de.  In  the  same  manner,  we  have  AF  =  af 
and  DF  =  df.  Therefore,  the  triangles  AEF  and  aef  are  equal 
(I.  76),  and  we  have  EF  =  ef.  Finally,  the  triangles  EDF  and  edf, 
being  mutually  equilateral,  are  equal;  therefore,  the  angle  EDF, 
which  measures  the  diedral  angle  SA,  is  equal  to  the  angle  edf,  which 
measures  the  diedral  angle  sa,  and  the  diedral  angles  SA  and  sa  are 
equal  (41).  In  the  same  manner,  it  may  be  proved  that  the  diedral 
angles  SB  and  SC  are  equal  to  the  diedral  angles  sb  and  sc,  re 
spectively. 

So  far  the  demonstration  applies  to  either  of  the  two  figures 
denoted  in  the  diagram  by  s-abc,  which  are  symmetrical  with  each 
other.  If  the  first  of  these  figures  is  given,  it  follows  that  S  and  s 
are  equal,  since  they  can  evidently  be  applied  to  each  other  so  as  to 
coincide  in  all  their  parts  (66)  ;  if  the  second  is  given,  it  follows  that 
S  and  s  are  symmetrical  (68). 


BOOK   VII. 

POLYEDEONS. 

1.  DEFINITION.  A  polyedron  is  a  geometrical  solid  bounded  by 
planes. 

The  bounding  planes,  by  their  mutual  intersections,  limit  each 
other,  and  determine  the  faces  (which  are  polygons),  the  edges,  and 
the  vertices,  of  the  polyedron.  A  diagonal  of  a  polyedron  is  a 
straight  line  joining  any  two  of  its  vertices  not  in  the  same  face. 

The  least'  number  of  planes  that  can  form  a  polyedral  angle  is 
three ;  but  the  space  within  the  angle  is  indefinite  in  extent,  and  it 
requires  a  fourth  plane  to  enclose  a  finite  portion  of  space,  or  to  form 
a  solid  ;  hence,  the  least  number  of  planes  that  can  form  a  polyedron 
is  four. 

2.  Definition.  A  polyedron  of  four  faces  is  called  a  tetraedron; 
one  of  six  faces,  a  hexaedron;   one  of  eight  faces,  an  octaedron; 
one  of  twelve  faces,  a  dodecaedron;  one  of  twenty  faces,  an  icosa- 
edron. 

3.  Definition.  A  polyedron  is  convex  when  the  sec'tion,  formed  by 
any  plane  intersecting  it,  is  a  convex  polygon. 

All  the  polyedrons  treated  of  in  this  work  will  be  understood  to 
be  convex. 

4.  Definition.    The   volume   of  any  polyedron   is   the   numerical 
measure   of   its   magnitude,  referred   to   some  other   polyedron   as 
the  unit.     The  polyedron  adopted  as  the  unit  is  called  the  unit  of 
volume. 

To  measure  the  volume  of  a  polyedron  is,  then,  to  find  its  ratio  to 

the  unit  of  volume. 

5.  Definition.  Equivalent    solids    are    those    which    have    equal 
volumes. 

196 


BOOK     VII. 


197 


PRISMS    AND    PARALLELOPIPEDS. 

6.  Definitions.  A  prism  is  a  polyedron  two  of 
whose  faces  are  equal  polygons  lying  in  parallel 
planes  and  having  their  homologous  sides  parallel, 
the  other  faces  being  parallelograms  formed  by 
the   intersections   of   planes   passed   through   the 
homologous  sides  of  the  equal  polygons. 

The  parallel  faces  are  called  the  bases  of  the 
prism;   the  parallelograms  taken  together  constitute  its  lateral  or 
convex  surface;  the  intersections  of  the  lateral  faces  are  its  lateral 
edges. 

The  altitude  of  a  prism  is  the  perpendicular  distance  between  the 
planes  of  its  bases. 

A  triangular  prism  is  one  whose  base  is  a  triangle ;  a  quadrangular 
prism,  one  whose  base  is  a  quadrilateral ;  etc. 

7.  Definitions.   A   right  prism  is  one  whose  lateral 
edges  are  perpendicular  to  the  planes  of  its  bases. 

In  a  right  prism,  any  lateral  edge  is  equal  to  the 
altitude. 

An  oblique  prism  is  one  whose  lateral  edges  are  ob 
lique  to  the  planes  of  its  bases. 


\ 


In  an  oblique  prism,  a  lateral  edge  is  greater  than  the  altitude. 
8.  Definition.  A  regular  prism  is  a  right  prism  whose  bases  are 
regular  polygons. 


9.  Definition.  If  a  prism,  ABCDE-F,  is 
intersected  by  a  plane  GK,  not  parallel 
to  its  base,  the  portion  of  the  prism  in 
cluded  between  the  base  and  this  plane, 
namely  ABCDE-GHIKL,  is  called  a 
truncated  prism. 


10.  Definition.  If  a  plane  intersects  a  prism  at  right  angles  to  its 
lateral  edges,  the  section  is  called  a  right  section  of  the  prism. 
17  * 


198 


GEOMETRY. 


11.  Definition.    A    parallelepiped    is    a    prism 
whose  bases  are  parallelograms.     It  is  therefore 
a  polyedron   all   of   whose    faces    are  parallelo 
grams. 

From  this  definition  and  (VI.  32)  it  is  evident 
that  any  two  opposite  faces  of  a  parallelepiped 
are  equal  parallelograms. 

12.  Definition.  A  right  parallelopiped  is  a  parallele 
piped  whose  lateral  edges  are  perpendicular  to  the 
planes  of  its  bases.     Hence,  by  (VI.  6),  its  lateral 
faces   are   rectangles;   but   its   bases   may  be  either 
rhomboids  or  rectangles. 

A  rectangular  parallelopiped  is  a  right  parallelopiped 
whose  bases  are  rectangles.     Hence  it  is  a  parallelopiped  all  of 
whose  faces  are  rectangles. 

Since  the  perspective  of  figures  in  space  distorts  the  angles,  the 
diagram  may  represent  either  a  right,  or  a  rectangular,  parallel 
opiped. 


13.  Definition.  A  cube  is  a  rectangular  parallelopiped 
whose  six  faces  are  all  squares. 


PROPOSITION  I.— THEOREM. 

14.   The  sections  of  a  prism  made  by  parallel  planes  are  equal 
polygons. 

Let  the  prism  AD '  be  intersected  by  the 
parallel  planes  GK,  G'K'-,  then,  the  sec 
tions,  GHIKL,  G'HTK'L',  are  equal 
polygons. 

For,  the  sides  of  these  polygons  are  paral 
lel,  each  to  each,  as  for  example,  GH  and 
G'H',  being  the  intersections  of  parallel 
planes  with  a  third  plane  (VI.  25),  and 
they  are  equal,  being  parallels  included 
between  parallels  (I.  104) ;  hence,  also,  the 
angles  of  the  polygons  are  equal,  each  to 


Q'l' 


BOOK     VII.  199 

each  (VI.  32).     Therefore,  the  two  sections,  being  both  mutually 
equilateral  and  mutually  equiangular,  are  equal. 

15.  Corollary.  Any  section  of  a  prism,  made  by  a  plane  parallel 
to  the  base,  is  equal  to  the  base. 

PROPOSITION  II.— THEOEEM. 

16.  The  lateral  area  of  a  prism  is  equal  to  the  product  of  the  perim 
eter  of  a  right  section  of  the  prism  by  a  lateral  edge. 

Let  AD'  be  a  prism,  and  GHIKL  a  right 
section  of  it ;  then,  the  area  of  the  convex  sur 
face  of  the  prism  is  equal  to  the  perimeter 
GHIKL  multiplied  by  a  lateral  edge  A  A ' . 

For,,  the  sides  of  the  section  GHIKL  being 
perpendicular  to  the  lateral  edges  A  A', 
BB ',  etc.,  are  the  altitudes  of  the  parallelo 
grams  which  form  the  convex  surface  of  the 
prism,  if  we  take  as  the  bases  of  these  paral 
lelograms  the  lateral  edges,  AA'  BB',  etc.,  which  are  all  equal. 
Hence,  the  area  of  the  sum  of  these  parallelograms  is  (IV.  10), 

GH  X  AA'  -f  HI  X  BB'  +  etc. 
=  (GH  +  HI  +  etc.)  X  AA'. 

17.  Corollary.  The  lateral  area  of  a  right  prism  is  equal  to  the 
product  of  the  perimeter  of  its  base  by  its  altitude. 

PROPOSITION  III.— THEOREM. 

18.  The  four  diagonals  of  a  parallelopiped  bisect  each  other. 

Let  ABCD-G  be  a  parallelopiped;  its  four  diagonals,  AG,  EC, 
BH,  DF,  bisect  each  other. 

Through  the  opposite  and  parallel  edges 
AE,  CG,  pass  a  plane  which  intersects  the 
parallel  faces  AS  CD,  EFGH,  in  the  parallel 
lines  A  C  and  EG.  The  figure  A  CGE  is  a 
parallelogram,  and  its  diagonals  A  G  and  EC 
bisect  each  other  in  the  point  0.  In  the  \~~4~~f~~~S\\ 
same  manner  it  is  shown  that  A  G  and  BH,  \  f//*"*^  \\ 

A  G  and  DF,  biseQt  each  other ;  therefore,  the 
four  diagonals  bisect  each  other  in  the  point  0. 


200  GEOMETRY. 

19.  Scholium.  The  point  0,  in  which  the  four  diagonals  intersect, 
is  called  the  centre  of  the  parallelopiped ;  and  it  is  easily  proved  that 
any  straight  line  drawn  through  0  and  terminated  by  two  opposite 
faces  of  the  parallelopiped  is  bisected  in  that  point. 


PROPOSITION  IV.—  THEOREM. 

20.   The  sum  of  the  squares  of  the  four  diagonals  of  a  parallelopiped 
is  equal  to  the  sum  of  the  squares  of  its  twelve  edges. 
In  the  parallelogram  A  CGE  we  have  (III.  64), 


CE*  =  2AE~2  +  2ZC2, 
and  in  the  parallelogram  DBFH, 


BH2  +  DF2  =  2BF    +  2BD\ 

Adding,  and  observing  that  BF  =  AE, 
and  also  that  in  the  parallelogram  ABCD, 
2AC2  -f-  2BD2  =  4ZZP  +  4AD  2,  we  have 


AG2  +  CE*  +  BH2  +  DF  =  4          +  4B  2  +  4D, 

which  proves  the  theorem. 

21.  Corollary.  In  a  rectangular  parallelopiped,  the  four  diagonals 
are  equal  to  each  other  ;  and  the  square  of  a  diagonal  is  equal  to  the 
sum  of  the  squares  of  the  three  edges  which  meet  at  a  common  vertex. 
Thus,  if  A  G  is  a  rectangular  parallelopiped,  we  have,  by  dividing 
the  preceding  equation  by  4, 

AG2  =  AE2  +  Z52  +  AD2. 


22.  Scholium.  If  any  three  straight  lines  AB,  AE,  AD,  not  in  the 
same  plane,  are  given,  meeting  in  a  common  point,  a  parallelopiped 
can  be  constructed  upon  them.  For,  pass  a  plane  through  the 
extremity  of  each  line  parallel  to  the  plane  of  the  other  two ;  these 
planes,  together  with  the  planes  of  the  given  lines,  determine  the 
parallelopiped. 

In  a  rectangular  parallelopiped,  if  the  plane  of  two  of  the  three 
edges  which  meet  at  a  common  vertex  is  taken  as  a  base,  the  third 
edge  is  the  altitude.  These  three  edges,  or  the  three  perpendicular 


BOOK     VII. 


201 


distances  between  the  opposite  faces  of  a  rectangular  parallelepiped, 
are  called  its  three  dimensions. 


PKOPOSITION  V.— THEOEEM. 

23.  Two  prisms  are  equal,  if  three  faces  including  a  triedral  angle  of 
the  one  are  respectively  equal  to  three  faces  similarly  placed  including  a 
triedral  angle  of  the  other. 

Let  the  triedral  angles  A  and 
a  of  the  prisms  ABODE- A' , 
abcde-ar,  be  contained  by  equal 
faces  similarly  placed,  namely, 
ABCDE  equal  to  abcde,  AB' 
equal  to  a b',  and  AEr  equal  to 
ae' ;  then,  the  prisms  are  equal. 

For,  the  triedral  angles  A  and 

a  are  equal  (VI.  71),  and  can  be  applied,  the  one  to  the  other,  so  as 
to  coincide;  and  then  the  bases  ABCDE,  abcde,  coinciding,  the  face 
AB'  will  coincide  with  ab',  and  the  face  AE'  with  ae'  \  therefore 
the  sides  A'B',  A'E',  of  the  upper  base  of  one  prism,  will  coincide 
with  the  sides  a'b',  a'e't  of  the  upper  base  of  the  other  prism,  and 
since  these  bases  are  equal  they  will  coincide  throughout;  conag- 
quently  also  the  lateral  faces  of  the  two  prisms  will  coincide,  each 
to  each,  and  the  prisms  will  coincide  throughout ;  therefore,  the  prisms 
are  equal. 

24.  Corollary  I.   Two  truncated  prisms  are  equal,  if  three  faces  in 
cluding  a  triedral  angle  of  the  one  are  respectively  equal  to  three  faces 
similarly  placed  including  a  triedral  angle  of  the  other.     For,  the  pre 
ceding  demonstration  applies  whether  the  planes  A'B'C'D'E'  and 
a'b'c'd'e'  are  parallel  or  inclined  to  the  lower  bases. 

25.  Corollary  II.   Two  right  prisms  are  equal,  if  they  have  equal 
bases  and  equal  altitudes. 

In  the  case  of  right  prisms,  it  is  not 
necessary  to  add  the  condition  that 
the  faces  shall  be  similarly  placed ; 
for,  if  the  two  right  prisms  ABC-A' 
abc-a',  cannot  be  made  to  coincide  by 
placing  the  base  ABC  upon  the  equal 
base  abe ;  yet,  by  inverting  one  of  the 


202 


GEOMETRY. 


prisms  and  applying  the  base  ABC  to  the  base  a'b'e',  they  will 
coincide. 

PROPOSITION  VI.— THEOREM. 

26.  Any  oblique  prism  is  equivalent  to  a  right  prism  whose  base  is  a 
right  section  of  the  oblique  prism,  and  whose  altitude  is  equal  to  a  lateral 
edge  of  the  oblique  prism. 

Let  AB  CDE-A'  be  the  oblique  prism.  At 
any  point  F  in  the  edge  AA',  pass  a  plane 
perpendicular  to  AA'  and  forming  the  right 
section  FGHIK.  Produce  AA'  to  F',  mak 
ing  FF'=AA't  and  through  F'  pass  a 
second  plane  perpendicular  to  the  edge 
AA',  intersecting  all  the  faces  of  the 
prism  produced,  and  forming  another  right 
section  F'  G  'H'l'K'.  parallel  and  equal  to 
the  first.  The  prism  FGHIK-F'  is  a  right 
prism  whose  base  is  the  right  section  and 

whose  altitude  FF'  is   equal   to   the   lateral   edge  of  the  oblique 
prism. 

a  The  solid  ABCDE-Fis  a  truncated  prism  which  is  equal  to  the 
truncated  prism  A'B'C'D'E '-F'  (24).  Taking  the  first  away  from 
the  whole  solid  ABCDE-F',  there  remains  the  right  prism;  taking 
the  second  away  from  the  same  solid,  there  remains  the  oblique 
prism ;  therefore,  the  right  prism  and  the  oblique  prism  have  the 
same  volume,  that  is,  they  are  equivalent. 

r< 

PROPOSITION  VII.— THEOREM. 

27.  The  plane  passed  through  two  diagonally  opposite  edges  of  a 
parallelopiped  divides  it  into  two  equivalent  triangular  prisms. 

Let  ABCD-A  be  any  parallelopiped  ;  the 
plane  AGC'A',  passed  through  its  opposite 
edges  AA  and  CCf,  divides  it  into  two  equiv 
alent  triangular  prisms  ABG-A  and  A^G-A'. 

Let  FGHI  be  any  right  section  of  the 
parallelopiped,  made  by  a  plane  perpendicu 
lar  to  the  edge  AA.  The  intersection,  FH, 
of  this  plane  with  the  plane  A  G',  is  the  di- 


BOOK     VII. 


203 


agonal  of  the  parallelogram  FGHI,  and  divides  that  parallelogram 
into  two  equal  triangles,  FGH  and  FIH.  The  oblique  prism  AB  C-A 
is  equivalent  to  a  right  prism  whose  base  is  the  triangle  FGH  and 
whose  altitude  is  AA  (26)  ;  and  the  oblique  prism  AD  C-A  is  equiva 
lent  to  a  right  prism  whose  base  is  the  triangle  FIH  and  whose 
altitude  is  A  A.  The  two  right  prisms  are  equal  (25) ;  therefore, 
the  oblique  prisms,  which  are  respectively  equivalent  to  them,  are 
equivalent  to  each  other. 

PROPOSITION  VIII.— THEOREM. 

28.   Two  rectangular  parallelopipeds  having  equal  bases  are  to  each 
other  as  their  altitudes. 

Let  P  and  Q  be  two  rectangular  par 
allelopipeds  having  equal  bases,  and  let 
AB  and  CD  be  their  altitudes. 

1st.  Suppose  the  altitudes  have  a  com 
mon  measure,  which  is  contained,  for 
example,  5  times  in  AB  and  3  times  in 
CD,  so  that  if  AB  is  divided  in  5  equal 
parts,  CD  will  contain  3  of  these  parts ; 
then  we  have 


\*     \ 

\ 

Q 

\ 

\ 

D            \ 

\ 

\ 

\ 

\ 

\ 

\ 

\L  _i 

\ 

\ 

CD      8 

If  now  we  pass  planes  through  the  several  points  of  division  of  AB 
and  CD,  perpendicular  to  these  lines,  the  parallelopiped  P  will  be 
divided  into  5  equal  parallelopipeds,  and  Q  into  3  parallelopipeds, 
each  equal  to  those  in  P  ;  hence, 


and,  therefore, 

P      AB 

Q~~  CD 

2d.  If  the  altitudes  are  incommensurable,  the  proof  may  be  given 
by  the  method  exemplified  in  (II.  51)  and  (III.  15),  or,  according  to 
the  method  of  limits,  as  follows. 

Let  CD  be  divided  into  any  number  of  equal  parts,  and  let  one 
of  these  parts  be  applied  to  AB  as  often  as  AB  will  contain  it. 


204 


GEOMETRY. 


Since  AB  and  CD  are  incommensurable, 
a  certain  number  of  these  parts  will 
extend  from  A  to  a  point  B',  leaving  a 
remainder  BB'  less  than  one  of  the  parts. 
Through  B'  pass  a  plane  perpendicular 
to  AB,  and  denote  the  parallelepiped 
whose  base  is  the  same  as  that  of  P  or 
Q,  and  whose  altitude  is  AB',  by  P'; 
then,  since  AB'  and  CD  are  commensur 
able, 

P'       AB' 

Q  ~~  CD' 

Now,  suppose  the  number  of  parts  into  which  CD  is  divided  to  be 
continually  increased ;  the  length  of  each  part  will  become  less  and 
less,  and  the  point  B'  will  approach  nearer  and  nearer  to  J5.  The 
limit  of  AB'  will  be  AB,  and  the  limit  of  P'  will  be  P  (V.  28). 

Pr  P  AB'  AB 

The  limit  of  —  will  therefore  be  -»  and  that  of  — —  will  be  — • 

pi  AB' 

Since,  then,  the  variables  —   and   are  constantly  equal  and 

Q  CD 

approach  two  limits,  these  limits  are  equal  (V.  29),  and  we  have 


P 

Q 


AB 
CD 


29.  Scholium.  The  three  edges  of  a  rectangular  parallelepiped 
which  meet  at  a  common  vertex  being  called  its  dimensions,  the  pre 
ceding  theorem  may  also  be  expressed  as  follows : 

Two  rectangular  parallelepipeds  which  have  two  dimensions  in  com 
mon  are  to  each  other  as  their  third  dimensions. 


PROPOSITION  IX.— THEOREM. 

30.  Two  rectangular  parallelopipeds  having  equal  altitudes  are  to 
each  other  as  their  bases. 

Let  a,  b  and  c  be  the  three  dimensions  of  the  rectangular  par 
allelepiped  P\  m,n  and  c  those  of  the  rectangular  parallelepiped  §; 
the  dimension  c,  or  the  altitude,  being  common. 


BOOK      VII. 


205 


Let  R  be  a  third  rectangular  parallel 
epiped  whose  dimensions  are  m,  b  and  c; 
then,  R  has  the  two  dimensions  b  and  c  in 
common  with  P,  and  the  two  dimensions 
m  and  c  in  common  with  Q ;  hence  (29), 


P 
R 


R 


and  multiplying  these  ratios  together, 

P  =  a  X  b 
Q~~  m  X  n 

But  a  X  b  is  the  area  of  the  base  of  P,  and  m  X  n  is  the  area  of 
the  base  of  Q ;  therefore,  P  and  Q  are  in  the  ratio  of  their  bases. 
31.  Scholium.  This  proposition  may  also  be  expressed  as  follows: 
Two  rectangular  parallelopipeds  which  have  one  dimension  in  com 
mon,  are  to  each  other  in  the  products  of  the  other  two  dimensions. 


PROPOSITION  X.— THEOREM. 

32.  Any  two  rectangular  parallelopipeds  are  to  each  other  as  the  pro 
ducts  of  their  three  dimensions. 

Let  a,  b  and  c  be  the  three  dimensions 
of  the  rectangular  parallelepiped  P; 
m,  n  and  p  those  of  the  rectangular 
parallelepiped  Q. 

Let  R  be  a  third  rectangular  paral 
lelepiped  whose  dimensions  are  a,  b  and 
p ;  then  R  has  two  dimensions  in  com 
mon  with  P  and  one  dimension  in  com 
mon  with  §;  hence,  by  (29)  and  (31), 


p 

\ 

\ 

\      \ 

p 

k  \ 

\ 

\ 

P 

R 


R 


p          y        mXn 
and  multiplying  these  ratios  together, 

P=  a  X  b  X  e 

Q  ~  m  X  n  X  p 

18 


\ 


206 


GEOMETRY. 


PROPOSITION  XI.— THEOREM. 

33.  The  volume  of  a  rectangular  parallelopiped  is  equal  to  the  pro 
duct  of  its  three  dimensions,  the  unit  of  volume  being  the  cube  whose 
edge  is  the  linear  unit. 

Let  a,  b,  c,  be  the  three  dimensions 
of  the  rectangular  parallelopiped  P; 
and  let  Q  be  the  cube  whose  edge  is  the 
linear  unit.  The  three  dimensions  of  Q 
are  each  equal  to  unity,  and  we  have, 
by  the  preceding  proposition. 

P       a  X  b  X  c 


1  X  1  XI 


=  a  X  b  X  c. 


P. 


Now,  Q  being  taken  as  the  unit  of  volume,  —  is  the  numerical  mea- 

Q 
sure,  or  volume  of  P,  in  terms  of  this  unit  (4) ;  therefore  the  volume 

of  P  is  equal  to  the  product  a  X  b  X  c. 

34.  Scholium  I.  Since  the  product  a  X  b  represents  the  base,  when 
c  is  called  the  altitude,  of  the  parallelopiped,  this  proposition  may 

•  also  be  expressed  as  follows : 

The  volume  of  a  rectangular  parallelopiped  is  equal  to  the  product 
of  its  base  by  its  altitude. 

35.  Sclwlium  II.  When  the  three  dimensions  of  the  parallelopiped 
are  each  exactly  divisible  by  the  linear  unit,  the  truth  of  the  propo 
sition  is  rendered  evident  by  dividing  the  solid 

into  cubes,  each  of  which  is  equal  to  the  unit  of 

volume.     Thus,  if  the  three  edges  which  meet  at 

a  common  vertex  A  are,  respectively,  equal  to  3, 

4  and  5,  times  the  linear  unit,  these  edges  may 

be  divided  respectively  into  3,  4  and  5  equal 

parts,  and  then  planes  passed  through  the  several 

points  of  division  at  right  angles  to  these  edges 

will  divide  the  solid  into  cubes,  each  equal  to  the  unit  cube,  the 

number  of  which  is  evidently  3x4x5. 

But  the  more  general  demonstration,  above  given,  includes  also 
the  cases  in  which  one  of  the  dimensions,  or  two  of  them,  or  all  three, 
are  incommensurable  with  the  linear  unit. 


BOOK     VII. 


207 


36.  Scholium  III.  If  the  three  dimensions  of  a  rectangular  paral 
lelepiped  are  each  equal  to  a,  the  solid  is  a  cube  whose  edge  is  a,  and 
its  volume  is  a  X  «  X  <*>  =  «3 1  or,  the  volume  of  a  cube  is  the  third 
power  of  its  edge.  Hence  it  is  that  in  arithmetic  and  algebra,  the 
expression  "  cube  of  a  number"  has  been  adopted  to  signify  the 
"  third  power  of  a  number." 


PKOPOSITION  XII.— THEOEEM. 

37.  The  volume  of  any  parallelopiped  is  equal  to  the  product  of  its 
base  by  its  altitude. 

Let  ABCD-A'  be  any  oblique  parallelopiped,  whose  base  is 
ABCD,  and  altitude  Br  0. 


D' 


Produce  the  edges  AB,  A'B',  DC,  D'C'',  in  AB  produced  take 
FG=AB,  and  through  .Fand  G  pass  planes,  FFTI,  GG'H'H, 
perpendicular  to  the  produced  edges,  forming  the  right  parallelopiped 
FGHI-F',vfith  the  base  FF'I'I  and  altitude  FG,  equivalent  to  the 
given  oblique  parallelopiped  ABCD-A'  (26). 

From  F',  draw  FK  perpendicular  to  FI  or  FT.  Since  AF  is 
perpendicular  to  the  plane  FI',  the  plane  of  the  base  and  the  plane 
FI'  are  perpendicular  to  each  other  (VI.  47) ;  therefore,  F'K  is 
perpendicular  to  the  plane  of  the  base  (VI.  49)  and  is  equal  to  B'O. 

Now  the  three  lines  F'G',  FT  and  F'K  are  perpendicular  to 
each  other ;  consequently  the  parallelopiped  KLMN-F',  constructed 
upon  them,  is  rectangular.  The  parallelopiped  FGHI-F',  regarded 
as  an  oblique  prism  whose  base  is  FGG'F'  and  lateral  edge  F'l', 
is  equivalent  to  the  right  prism,  or  rectangular  parallelopiped, 
KLMN-F',  whose  base  is  the  right  section  F'L  and  whose  altitude 


208 


GEOMETRY. 


is  FT  (26).  Therefore,  the  given  parallelepiped  ABCD-A'  is  also 
equivalent  to  the  rectangular  parallelepiped  KLMN-F'.  The  volume 
of  this  rectangular  parallelepiped  is  equal  to  the  product  of  its  base 
KM  by  its  altitude  F'K\  its  base  KM  is  equal  to  F'H',  or  FH, 
which  is  equivalent  to  AC,  and  its  altitude  F'K  is  equal  to  B'O; 
therefore  the  volume  of  the  parallelepiped  ABCD-A'  is  equal  to  the 
product  of  its  base  A  Cloy  its  altitude  B' 0. 


PROPOSITION  XIII.— THEOREM. 

38.  The  volume  of  any  prism  is  equal  to  the  product  of  its  base  by  its 
altitude. 

1st.  Let  ABC-A  be  a  triangular  prism. 
This  prism  is  equivalent  to  one-half  the  par- 
allelopiped  ABCD—A'  constructed  upon  the 
edges  AB,  BC  and  BB'  (27),  and  it  has  the 
same  altitude.  The  volume  of  the  parallele 
piped  is  equal  to  its  base  BD  multiplied  by  its 
altitude ;  therefore,  the  volume  of  the  triangu 
lar  prism  is  equal  to  its  base  ABC,  the  half  of  BD,  multiplied  by 
its  altitude. 

2d.  Let  ABODE- A'  be  any  prism.  It  may 
be  divided  into  triangular  prisms  by  planes 
passed  through  a  lateral  edge  AA  and  the  sev 
eral  diagonals  of  its  base.  The  volume  of  the 
given  prism  is  the  sum  of  the  volumes  of  the 
triangular  prisms,  or  the  sum  of  their  bases 
multiplied  by  their  common  altitude,  which  is 
the  base  AB  CDE  of  the  given  prism  multiplied  by  its  altitude. 

39.  Corollary.  Prisms  having  equivalent  bases  are  to  each  other  as 
their  altitudes ;  prisms  having  equal  altitudes  are  to  each  other  as 
their  bases ;  and  any  two  prisms  are  to  each  other  as  the  products 
of  their  bases  and  altitudes. 


BOOK    VII.  209 


PYKAMIDS. 

40.  Definitions.  A  pyramid  is  a  polyedron  bounded  by  a  polygon 
and  triangular  faces  formed  by  the  intersections 

of  planes  passed  through  the  sides  of  the  poly 
gon  and  a  common  point  out  of  its  plane;  as 
S-ABCDE. 

The  polygon,  ABODE,  is  the  base  of  the  pyra 
mid  ;  the  point,  S,  in  which  the  triangular  faces 
meet,  is  its  vertex ;  the  triangular  faces  taken  to 
gether  constitute  its  lateral,  or  convex,  surface ;  the 
area  of  this  surface  is  the  lateral  area;  the  lines 
SA,  SB,  etc.,  in  which  the  lateral  faces  intersect,are 
its  lateral  edges.  The  altitude  of  the  pyramid  is  the  perpendicular 
distance  SO  from  the  vertex  to  the  base. 

A  triangular  pyramid  is  one  whose  base  is  a  triangle;  a  quadrangu 
lar  pyramid,  one  whose  base  is  a  quadrilateral ;  etc. 

A  triangular  pyramid,  having  but  four  faces  (all  of  which  are 
triangles),  is  a  tetraedron ;  and  any  one  of  its  faces  may  be  taken  as 
its  base. 

41.  Definitions.  A  regular  pyramid  is  one  whose  base  is  a  regular 
polygon,   and  whose  vertex  is  in  the  perpendicular 

to  the  base  erected  at  the  centre  of  the  polygon. 
This  perpendicular  is  called  the  axis  of  the  regular 
pyramid. 

From  this  definition  and  (VI.  10)  it  follows  that 
all  the  lateral  faces  of  a  regular  pyramid  are  equal 
isosceles  triangles. 

The  slant  height  of  a  regular  pyramid  is  the  per 
pendicular  from  the  vertex  to  the  base  of  any  one 
of  its  lateral  faces. 

42.  Definitions.    A  truncated  pyramid  is  the  portion  of  a  pyra 
mid  included  between  its  base  and  a  plane  cutting  all  its  lateral 
edges. 

When  the  cutting  plane  is  parallel  to  the  base,  the  truncated  pyra 
mid  is  called  a  frustum  of  a  pyramid.  The  altitude  of  a  frustum  is 
the  perpendicular  distance  between  its  bases. 

18*  O 


210 


GEOMETRY. 


In  a  frustum  of  a  regular  pyramid,  the  lateral  faces  are  equal 
trapezoids ;  and  the  perpendicular  distance  between  the  parallel 
sides  of  any  one  of  these  trapezoids  is  the  slant  height  of  the 
frustum. 


PKOPOSITION  XIV— THEOEEM. 

43.  If  a  pyramid  is  cut  by  a  plane  parallel  to  its  base :  1st,  the  edges 
and  the  altitude  are  divided  proportionally ;  2d,  the  section  is  a  polygon 
similar  to  the  base. 

Let  the  pyramid  S- ABCDE,  whose  altitude 
is  SO,  be  cut  by  the  plane  abcde  parallel  to  the 
base,  intersecting  the  lateral  edges  in  the  points 
a,  b,  c,  d,  e,  and  the  altitude  in  o ;  then, 

1st.  The  edges  and  the  altitude  are  divided 
proportionally. 

For,  suppose  a  plane  passed  through  the  ver 
tex  S  parallel  to  the  base ;  then,  the  edges  and 
altitude,  being  intersected  by  three  parallel 
planes,  are  divided  proportionally  (VI.  37),  and 
we  have 


Sa 

SA 


Sb_Se 
SB~  SC 


So 
SO 


2d.  The  section  abcde  is  similar  to  the  base  ABCDE. 

For,  the  sides  ab,  be,  etc.,  are  parallel  respectively  to  AB,  BC,  etc. 
(VI.  25),  and  in  the  same  directions :  therefore  the  angles  of  the  two 
polygons  are  equal,  each  to  each  (VI.  32). 

Also,  since  ab  is  parallel  to  AB,  and  be  parallel  to  BC,  the  tri 
angles  Sab  and  SAB  are  similar,  and  the  triangles  Sbc  and  SB  C  are 
similar;  therefore, 


ab^    _  £&  bc_  _  £6 

AB~  SB'™    BC~  SB 


whence 


ab 
AB 


be 
BC' 


BOOK     VII. 


211 


and  in  the  same  manner  we  should  find 


EC 


cd 
CD 


de_ 
DE 


ea 
EA 


Therefore,  the  polygons  abcde  and  ABCDE  are  similar  (III.  24). 
44."  Corollary  I.  The  polygons  abcde  and  ABCDE  being  similar, 
their  surfaces  are  proportional  to  the  squares  of  their  homologous 
sides;  hence 

abcde  ab          Sa          So 

ABCDE  ~~  AB*  ~  ~SA2  ~  SO* ' 

that  is,  the  surface  of  any  section  of  a  pyramid  parallel  to  its  base  is 
proportional  to  the  square  of  its  distance  from  the  vertex. 

45.  Corollary  II.  If  two  pyramids,  S- ABCDE  and  S '-A'B'C'D', 
having  equal  altitudes  SO  and  S'  0',  are  cut  by  planes  parallel  to  their 
bases  and  at  equal  distances,  So  and  S'o',  from  their  vertices,  the 
sections  abcde  and  a'b'c'd'  will  be  proportional  to  the  bases. 

For,  by  the  preceding  corollary, 


and 


abdce       _  So2 
ABCDE~~SO* 


a'b'c'd1 


A'B'C'D' 


whence,    since  So  —  S'o'   and 
S0  =  S'O', 

abcde  a'b'c'd' 

ABCDE~  A'B'C'D' 


A' 


46.  Corollary  III.  If  two  pyramids  have  equal  altitudes  and  equiva 
lent  bases,  sections  made  by  planes  parallel  to  their  bases  and  at  equal 
distances  from  their  vertices  are  equivalent. 


212 


GEOMETRY. 


PROPOSITION  XV.— THEOREM. 

47.   The  lateral  area  of  a  regular  pyramid  is  equal  to  the  product  of 
the  perimeter  of  its  base  by  one-half  its  slant  height. 


For,  let  S-ABCDE  be  a  regular  pyra 
mid  ;  the  lateral  faces  SAB,  SA  C,  etc.,  be 
ing  equal  isosceles  triangles,  whose  bases  are 
the  sides  of  the  regular  polygon  ABODE  and 
whose  common  altitude  is  the  slant  height 
SH,  the  sum  of  their  areas,  or  the  lateral  area 
of  the  pyramid,  is  equal  to  the  sum  of  AB, 
BC,  etc.,  multiplied  by  ±SH  (IV.  13). 


48.  Corollary.  The  lateral  area  of  the  frustum  of  a  regular  pyramid 
is  equal  to  the  half  sum  of  the  perimeters  of  its  bases  multiplied  by  the 
slant  height  of  the  frustum.  For,  this  product  is  the  measure  of  the 
sum  of  the  areas  of  the  trapezoids  ABba,  BCcb,  etc.,  whose  common 
altitude  is  the  slant  height  hH  (IV.  17). 


PROPOSITION  XVI.— LEMMA. 

49.  A  series  of  prisms  may  be  inscribed  in  any  given  triangular 
pyramid  whose  total  volume  shall  differ  from  the  volume  of  the  pyramid 
by  less  than  any  assigned  volume.  r< 

Let  S-AB  C  be  the  given  triangular 
pyramid,  whose  altitude  is  AT.  Divide 
the  altitude  AT  into  any  number  of 
equal  parts  Ax,  xy,  etc.,  and  denote 
one  of  these  parts  by  h.  Through  the 
points  of  division  x,  y,  etc.,  pass  planes 
parallel  to  the  base,  cutting  from  the 
pyramid  the  sections  DEF,  GHI,  etc. 
Upon  the  triangles  DEF,  GHI,  etc., 
as  upper  bases,  construct  prisms  whose 
lateral  edges  are  parallel  to  SA,  and 
whose  altitudes  are  each  equal  to  h.  This  is  effected  by  passing 


BOOK    VII.  213 

planes  through  EF,  HI,  etc.,  parallel  to  SA.  There  will  thus  be 
formed  a  series  of  prisms  DEF-A,  GHI-D,  etc.,  inscribed  in  the 
pyramid. 

Again,  upon  the  triangles  ABC,  DEF,  GUI,  etc.,  as  lower  bases, 
construct  prisms  whose  lateral  edges  are  parallel  to  SA,  and  whose 
altitudes  are  each  equal  to  h.  This  also  is  effected  by  passing  planes 
through  BC,  EF,  HI,  etc.,  parallel  to  SA.  There  will  thus  be 
formed  a  series  of  prisms  ABC-D,  DEF—G,  etc.,  which  may  be  said 
to  be  circumscribed  about  the  pyramid. 

Now,  the  first  inscribed  prism  DEF-A  is  equivalent  to  the  second 
circumscribed  prism  DEF-G,  since  they  have  the  same  base  DEF  and 
equal  altitudes  (39) ;  the  second  inscribed  prism  GHI-D  is  equivalent 
to  the  third  circumscribed  prism  GHI-K;  and  so  on.  Therefore,  the 
sum  of  all  the  inscribed  prisms  differs  from  the  sum  of  all  the  cir 
cumscribed  prisms  only  by  the  first  circumscribed  prism  ABC-D. 
But  the  pyramid  is  greater  than  the  sum  of  the  inscribed  prisms  and 
less  than  the  sum  of  the  circumscribed  prisms ;  therefore,  the  differ 
ence  between  the  total  volume  of  the  inscribed  prisms  and  the  volume 
of  the  pyramid  is  less  than  the  volume  of  the  prism  ABC-D. 

The  volume  of  the  prism  ABC-D  may  be  made  as  small  as  we 
please,  or  less  than  any  assigned  volume,  by  dividing  the  altitude 
A  T  into  a  sufficiently  great  number  of  equal  parts ;  for,  if  the  as 
signed  volume  is  represented  by  a  prism  whose  base  is  ABC  and 
altitude  Aa,  we  have  only  to  divide  A  T  into  a  number  of  equal  parts 
each  less  than  Aa. 

Therefore,  the  difference  between  the  total  volume  of  the  inscribed 
prisms  and  the  volume  of  the  pyramid  may  be  made  less  than  any 
assigned  volume. 

50.  Corollary.  If  the  number  of  parts  into  which  the  altitude  is 
divided  is  increased  indefinitely,  the  difference  between  the  volume 
of  the  inscribed  prisms  and  that  of  the  pyramid  approaches  indefi 
nitely  to  zero ;  and  therefore  the  pyramid  is  the  limit  of  the  sum 
of  the  inscribed  prisms,  as  their  number  is  indefinitely  increased 
(V.  28). 


214 


GEOMETRY. 


PROPOSITION  XVII.— THEOREM. 

51.   Two  triangular  pyramids  having  equivalent  bases  and  equal  alti 
tudes  are  equivalent. 

Let  S-ABC  and  S'-A'B'  C'  be  two  triangular  pyramids  having 


B  B< 

equivalent  bases,  ABC,  A'B'C',  in  the  same  plane,  and  a  common 
altitude  AT. 

Divide  the  altitude  AT  into  a  number  of  equal  parts  Ax,  xy,  yz, 
etc.,  and  through  the  points  of  division  pass  planes  parallel  to  the 
plane  of  the  bases,  intersecting  the  two  pyramids.  In  the  pyramid 
S-AB  C  inscribe  a  series  of  prisms  whose  upper  bases  are  the  sections 
DEF,  GHI,  etc.,  and  in  the  pyramid  S'-A'B'C'  inscribe  a  series  of 
prisms  whose  upper  bases  are  the  sections  D'E'F',  G'H'I',  etc. 
Since  the  corresponding  sections  are  equivalent  (46),  the  correspond 
ing  prisms,  having  equivalent  bases  and  equal  altitudes,  are  equiva 
lent  (39) ;  therefore,  the  sum  of  the  prisms  inscribed?  in  the  pyramid 
S-ABC  is  equivalent  to  the  sum  of  the  prisms  inscribed  in  the  pyra 
mid  S'-A'B'C'  \  that  is,  if  we  denote  the  total  volumes  of  the  two 
series  of  prisms  by  Fand  F',  we  have 

V  =  V. 

Now  let  the  number  of  equal  parts  into  which  the  altitude  is 
divided  be  supposed  to  be  indefinitely  increased;  the  volume  F 
approaches  to  the  volume  of  the  pyramid  S-ABC  as  its  limit,  and 
the  volume  F'  approaches  to  the  volume  of  the  pyramid  S'-A'B'C' 
as  its  limit  (50).  Since,  then,  the  variables  F  and  V  are  always 
equal  to  each  other  and  approach  two  limits,  these  limits  are  equal 
(V.  29) ;  that  is,  the  volumes  of  the  pyramids  are  equal. 


BOOK      VII. 


215 


PEOPOSITION  XVIII.— THEOEEM. 

52.  A  triangular  pyramid  is  one-third  of  a  triangular  prism  of  the 
same  base  and  altitude. 

Let  S-ABC  be  a  triangular  pyramid.  Through 
one  edge  of  the  base,  as  A  C,  pass  a  plane  A  CDE 
parallel  to  the  opposite  lateral  edge  SB,  and  through 
S  pass  a  plane  SED  parallel  to  the  base ;  the  prism 
ABC-E  has  the  same  base  and  altitude  as  the  given 
pyramid,  and  we  are  to  prove  that  the  pyramid  is 
one-third  of  the  prism. 

Taking  away  the  pyramid  S-ABC  from  the  prism,  there  remains 
a  quadrangular  pyramid  whose  base  is  the  parallelogram  A  CDE  and 
vertex  S.  The  plane  SEC,  passed  through  SE  and  SC,  divides  this 
pyramid  into  two  triangular  pyramids,  S-AEC  and  S-ECD,  which 
are  equivalent  to  each  other,  since  their  triangular  bases  AEG  and 
ECD  are  the  halves  of  the  parallelogram  A  CDE,  and  their  common 
altitude  is  the  perpendicular  from  S  upon  the  plane  A  CDE  (51). 
The  pyramid  S-ECD  may  be  regarded  as  having  ESD  as  its  base 
and  its  vertex  at  (7;  therefore,  it  is  equivalent  to  the  pyramid 
S-AB  C  which  has  an  equivalent  base  and  the  same  altitude.  There 
fore,  the  three  pyramids  into  which  the  prism  is  divided  are  equiva 
lent  to  each  other,  and  the  given  pyramid  is  one-third  of  the  prism. 

53.  Corollary.   The  volume  of  a  triangular  pyramid  is  equal  to  one- 
third  of  the  product  of  its  base  by  its  altitude. 


PKOPOSITION  XIX.— THEOREM. 

54.   The  volume  of  any  pyramid  is  equal  to  one-third  of  the 
of  its  base  by  its  altitude. 

For,  any  pyramid,  S-ABCDE,  may  be  di 
vided  into  triangular  pyramids  by  passing  planes 
through  an  edge  SA  and  the  diagonals  AD,  A  C, 
etc.,  of  its  base.  The  bases  of  these  pyramids 
are  the  triangles  which  compose  the  base  of  the 
given  pyramid,  and  their  common  altitude  is  the 
altitude  SO  of  the  given  pyramid.  The  volume 
of  the  given  pyramid  is  equal  to  the  sum  of  the 


product 


216  GEOMETRY. 

volumes  of  the  triangular  pyramids,  which  is  one-third  of  the  sum 
of  their  bases  multiplied  by  their  common  altitude,  or  one-third  the 
product  of  the  base  ABCDEloy  the  altitude  SO. 

55.  Corollary.  Pyramids  having  equivalent  bases  are  to  each  other  as 
their  altitudes.     Pyramids  having  equal  altitudes  are  to  each  other  as 
their  bases.     Any  two  pyramids  are  to  each  other  as  the  products  of 
their  bases  and  altitudes. 

56.  Scholium.   The  volume  of  any  polyedron  may  be  found  by 
dividing  it  into  pyramids,  and  computing  the  volumes  of  these  pyra 
mids  separately.     The  division  may  be  effected  by  drawing  all  the 
diagonals  that  can  be  drawn  from  a  common  vertex  ;  the  bases  of 
the  pyramids  will  be  all  the  faces  of  the  polyedron  except  those 
which  meet  at  the  common  vertex.    Or,  a  point  may  be  taken  within 
the  polyedron  and  joined  to  all  the  vertices  ;    the  polyedron  will 
then  be  decomposed  into  pyramids  whose  bases  will  be  the  faces  of 
the  polyedron,  and  whose  common  vertex  will  be  the  point  taken 
within  it. 

PKOPOSITION  XX.—  THEOKEM. 

57.  Two  tetraedrons  which  have  a  triedral  angle  of  the  one  equal  to 
a  triedral  angle  of  the  other,  are  to  each  other  as  the  products  of  the 
three  edges  of  the  equal  triedral  angles. 

Let  ABCD,  AB'C'D',  be  the 
given  tetraedrons,  placed  with  their 
equal  triedral  angles  in  coincidence 
at  A.  From  D  and  D',  let  fall  DO 
and  D'  0'  perpendicular  to  the  face 
A  B  C.  Then,  taking  the  faces  AB  (7,  • 
AB'  C',  as  the  bases  of  the  triangu 

lar  pyramids  D-ABC,  D'-AB'C',  and  denoting  the  volumes  by  V 
and  V',  we  have  (55), 

_Z          ABC  X  DO          ABC        DO 
Vf~AB'C'XD'Of       AB'C'       D'O'' 

By  (IV.  22)  and  (III.  25),  we  have 

ABC         AB  X  AC  DO         AD 


AB'C'~AB'  X  AC''         D'O'       ADf 


BOOK    VII.  217 

therefore, 

V       AB   XAC   X  AD 
V'~  AB'  X  AC'  X  AD'' 

PKOPOSITION  XXI.—  THEOEEM. 

58.  A  frustum  of  a  triangular  pyramid  is  equivalent  to  the  sum  of 
three  pyramids  whose  common  altitude  is  the  altitude  of  the  frustum, 
and  whose  bases  are  the  lower  base,  the  upper  base,  and  a  mean  pro 
portional  between  the  bases,  of  the  frustum. 

Let  ABG-D  be  a  frustum  of  a  tri 
angular  pyramid,  formed  by  a  plane 
DEF  parallel  to  the  base  ABC. 

Through  the  vertices  A,  E  and  C, 
pass  a  plane  AEG',  and  through  the 
vertices  E,  D  and  C,  pass  a  plane  EDO, 
dividing  the  frustum  into  three  pyra 
mids.  For  brevity,  denote  the  pyramid 

E-ABCby  P,  the  pyramid  E-DFC  by  p,  and  the  pyramid  E-ADC 
by  §. 

The  pyramids  P  and  Q,  regarded  as  having  the  common  vertex 
C  and  their  bases  in  the  same  plane  BD,  have  a  common  altitude 
and  are  to  each  other  as  their  bases  AEB  and  AED  (55).  But  the 
triangles  AEB  and  AED,  having  a  common  altitude,  namely,  the 
altitude  of  the  trapezoid  ABED,  are  to  each  other  as  their  bases  AB 
and  DE;  hence  we  have 

P      AB 

q    DE 

The  pyramids  Q  and  p,  regarded  as  having  the  common  vertex 
E  and  their  bases  in  the  same  plane  AF,  have  a  common  altitude, 
and  are  to  each  other  as  their  bases  ADC  aud  DCF.  But  the  tri 
angles  ADC  and  DCF,  having  a  common  altitude,  namely,  the  alti 
tude  of  the  trapezoid  A  CFD,  are  to  each  other  as  their  bases  A  C 
and  DF;  hence  we  have 


p       DF 

.  Moreover,  the  section  DEF  being  similar  to  ABC  (43),  we  have 

19 


218  GEOMETRY. 

AB^AC 
DE  ~~  DF 
and  therefore 

P 

Q~ 

whence 

Q2  =  P  X  p, 

that  is  (III.  5),  the  pyramid  Q  is  a  mean  proportional  between  the 
pyramids  P  and  p. 

Now,  denote  the  lower  base  ABC  of  the  frustum  by  B,  its  upper 
base  by  b,  and  its  altitude  by  h.  The  pyramid  P,  regarded  as  having 
its  vertex  at  E,  has  the  altitude  h  and  the  base  B  ;  the  pyramid  p, 
regarded  as  having  its  vertex  at  C,  has  the  altitude  h  and  the  base 
b  ;  hence  (54), 


and 


=     h  X      B  X 


consequently,  Q  is  equivalent  to  a  pyramid  whose  altitude  is  h  and 
whose  base  is  a  mean  proportional  between  the  bases  B  and  b  ;  and 
since  the  given  frustum  is  the  sum  of  P,  p  and  Q,  the  proposition  is 
established. 

If  F  denotes  the  volume  of  the  frustum,  the  proposition  is  ex 
pressed  by  the  formula 


F=£fcX  B  +  ^hXb  +  th  X  VBX  b, 
or 

V=  ±h  (B  +  b  +  i/£~X~6). 

59.  Corollary.  A  frustum  of  any  pyramid  is  equivalent  to  the  sum  of 
three  pyramids  whose  common  altitude  is  the  altitude  of  the  frustum,  and 
whose  bases  are  the  lower  base,  the  tipper  base,  and  a  mean  proportional 
between  the  bases,  of  the  frustum. 

For,  let  ABCDE-F  be  a  frustum  of  any  pyramid  S-ABCDE. 
Let  S'-A'B'C'  be  a  triangular  pyramid,  having  the  same  altitude 
as  the  pyramid  S-ABCDE,  and  a  base  A'B'C'  equivalent  to  the 
base  ABCDE,  and  in  the  same  plane  with  it.  The  volumes  of  the 
two  pyramids  are  equivalent  (55).  Let  the  plane  of  the  upper  base 
of  the  given  frustum  be  produced  to  cut  the  triangular  pyramid. 


BOOK      VII. 


219 


The  section  F'GT  being  equivalent  to  the  section  FOHIK  (46), 
the  pyramid  S'-F'GT  is  equivalent  to  the  pyramid  S-FGHIK; 


and  taking  away  these  pyramids  from  the  whole  pyramids,  the  frus 
tums  that  remain  are  equivalent ;  therefore,  denoting  by  B  the  area 
of  ABODE  QT  of  A'B'C',  by  b  that  of  FGHIK  or  of  F'G'I',  and 
by  h  the  common  altitude  of  the  two  frustums,  we  have  for  the  vol 
ume  of  the  given  frustum  the  same  expression  as  for  that  of  the  tri 
angular  frustum ;  namely, 


TRUNCATED    TEIANGULAE    PEISM. 

PEOPOSITION  XXIL— THEOREM. 

60.  A  truncated  triangular  prism  is  equivalent  to  the  sum  of  three 
pyramids  whose  common  base  is  the  base  of  the  prism,  and  whose  vertices 
are  the  three  vertices  of  the  inclined  section. 

Let  ABC-DEF  be  a  truncated  triangular 
prism  whose  base  is  ABC  and  inclined  sec 
tion  DEF. 

Pass  the  planes  AEC  and  DEC,  dividing 
the  truncated  prism  into  the  three  pyramids, 
E-ABC,  E-ACD  and  E-CDF. 

The  first  of  these  pyramids,  E-ABC,  has 
the  base  ABC  and  the  vertex  E. 


220 


GEOMETRY. 


The  second  pyramid,  E-ACD,  is  equivalent  to  the  pyramid 
B-A  CD ;  for  they  have  the  same  base  A  CD,  and  the  same  altitude, 
since  their  vertices  E  and  B  are  in  the  line  EB  parallel  to  this  base. 
But  the  pyramid  B-A  CD  is  the  same  as  D-ABC;  that  is,  it  has  the 
base  AB  C  and  the  vertex  D. 

The  third  pyramid,  E-CDF,  is  equivalent  to  the  pyramid  B-A  CF; 
for  they  have  equivalent  bases  CDF  and  A  CF  in  the  same  plane, 
and  also  the  same  altitude,  since  their  vertices  E  and  B  are  in  the 
line  EB  parallel  to  that  plane.  But  the  pyramid  B-A  CF  is  the 
same  as  F-ABC;  that  is,  it  has  the  base  AB  C  and  the  vertex  jP. 

Therefore  the  truncated  prism  is  equivalent  to  three  pyramids 
whose  common  base  is  ABC  and  whose  vertices  are  E,  D  and  F. 

61.  Corollary  I.  The  volume  of  a  truncated  right  triangular  prism 
is  equal  to  the  product  of  its  base  by  one-third  the  sum  of  its  lateral 
edges.  For.  the  lateral  edges  AD,  BE,  CF,  being  perpendicular  to 
the  base,  are  the  altitudes  of  the  three  pyramids  p 

to  which  the  truncated  prism  has  been  proved  to 
be  equivalent ;  therefore,  the  volume  is 


ABC  X  \AD  +  ABC  X 


+  ABC  X 


or 


ABCX 


AD  +  BE  +  CF 


62.  Corollary  II.  The  volume  of  any  truncated  triangular  prism  is 
equal  to  the  product  of  its  right  section  by  one-third  the  sum  of  its  lateral 
edges.  For,  let  ABC-A'B' C'  be  any  trun 
cated  triangular  prism ;  the  right  section 
DEF  divides  it  into  two  truncated  right  -^ 

prisms  whose  volumes  are,  by  the  preced 
ing  corollary, 


and 


the  sum  of  which  is 

DEFX 


AA'  f  BB'  +  CC' 


BOOK     VII.  221 


SIMILAE    POLYEDEONS. 

63.  Definition.  Similar  polyedrons  are  those  which  are  bounded  by 
the  same  number  of  faces  similar  each  to  each  and  similarly  placed, 
and  which  have  their  homologous  polyedral  angles  equal. 

Parts  similarly  placed  in  two  similar  polyedrons,  whether  faces, 
lines,  or  angles,  are  homologous. 

64.  Corollary  I.  Since  homologous  edges  are  in  the  ratio  of  simili 
tude  of  the  polygons  of  which  they  are  homologous  sides  (III.  24), 
and  every  edge  belongs  to  two  faces,  in  each  polyedron,  it  follows 
that  the  ratio  of  similitude  of  any  two  homologous  faces  is  the  same 
as  that  of  any  other  two  homologous  faces,  and  this  ratio  may  be 
called  the  ratio  of  similitude  of  the  two  polyedrons. 

Therefore,  any  two  homologous  edges  of  two  similar  polyedrons  are 
in  the  ratio  of  similitude  of  the  polyedrons ;  or,  homologous  edges  are 
proportional  to  each  other. 

65.  Corollary  II.   The  ratio  of  the  surfaces  of  any  two  homologous 
faces  is  the  square  of  the  ratio  of  similitude  of  the  polyedrons  (IV.  24) ; 
or,  any  two  homologous  faces  are  to  each  other  as  the  squares  of  any  two 
homologous  edges. 

Hence,  by  the  theory  of  proportions  (III.  12),  the  entire  surfaces 
of  two  similar  polyedrons  are  to  each  other  as  the  squares  of  any  two 
homologous  edges. 

PROPOSITION  XXIII.— THEOREM. 

66.  If  a  tetraedron  is  cut  by  a  plane  parallel  to  one  of  its  faces,  the 
tetraedron  cut  off  is  similar  to  the  first. 

Let  the  tetraedron  ABCD  be  cut  by  the 
plane  B'C'D'  parallel  to  BCD',  then,  the 
tetraedrons  AB'C'D'  and  ABCD  are  simi 
lar. 

For,  since  the  edges  AB,  A  C,  AD,  are 
divided  proportionally  at  B',  C',  Df,  the 
face  AB'C'  is  similar  to  the  face  ABC, 
AC'D'  to  A  CD,  and  AB'D'  to  ABD-, 
also,  B'C'D'  is  similar  to  BCD  (43). 
Moreover,  the  homologous  triedral  angles, 
being  contained  by  equal  face  angles  simi- 

19* 


222  GEOMETRY. 

larly  placed,  are  equal,  each  to  each  (VI.  71).     Therefore,  by  the 
definition  (63),  the  tetraedrons  are  similar. 


PKOPOSITION  XXIV.— THEOEEM. 

67.  Two  tetraedrons  are  similar,  when  a  diedral  angle  of  the  one  is 
equal  to  a  diedral  angle  of  the  other,  and  the  faces  including  these  angles 
are  similar  each  to  each,  and  similarly  placed. 

Let  AS  CD,  A'B'C'D',  be 
two  tetraedrons  in  which  the 
diedral  angle  AB  is  equal  to  the 
diedral  angle  A'B',  and  the 
faces  ABC  and  ABD  are  res 
pectively  similar  to  the  faces 
A'B'C'  and  A'B'D';  then,  the 
tetraedrons  are  similar. 

The  triedral  angles  A  and  A'  are  equal,  since  they  may  evidently 
be  placed  with  their  vertices  in  coincidence  so  as  to  coincide  in  all 
their  parts.  Therefore,  the  angles  CAD  and  C'A'D'  are  equal.  The 
given  similar  faces  furnish  the  proportions 

AC         AB  AD         AB 

A^C'  ~  A'B''        A'D'  ~~  AB'' 

whence 

AC  =_  AD 
A'C'~  AD1' 

therefore,  the  faces  A  CD  and  A'C'D'  are  similar  (III.  32). 

In  like  manner  it  is  shown  that  the  triedral  angles  B  and  Br  are 
equal,  and  the  faces  BCD  and  B'C'D'  are  similar. 

Finally,  the  triedral  angles  C  and  C"  are  equal,  since  their  face 
angles  are  equal  each  to  each  and  are  similarly  placed  (VI.  71) ; 
and  the  triedral  angles  D  and  D'  are  equal  for  the  same  reason. 
Therefore,  the  two  tetraedrons  are  similar  (63). 


BOOK    VII. 


223 


PEOPOSITION  XXV.— THEOKEM. 

68.  Two  similar  polyedrons  may  be  decomposed  into  the  same  number 
of  tetraedrons  similar  each  to  each  and  similarly  placed. 

Let  ABCDEFOH  and  abcdefyh  be  similar  polyedrons,  of  which 
A  and  a  are  homologous  vertices. 


Let  all  the  faces  not  adjacent  to  A,  in  the  first  polyedron,  be 
decomposed  into  triangles,  and  let  straight  lines  be  drawn  from  A  to 
the  vertices  of  these  triangles ;  the  polyedron  is  then  divided  into 
tetraedrons  having  these  triangles  as  bases  and  the  common  ver 
tex  A. 

Also  decompose  the  faces  not  adjacent  to  a,  in  the  second  polye 
dron,  into  triangles  similar  to  those  in  the  first  polyedron  and  simi 
larly  placed  (III.  39),  and  let  straight  lines  be  drawn  from  a  to  the 
vertices  of  these  triangles ;  the  second  polyedron  is  then  divided  into 
the  same  number  of  tetraedrons  as  the  first,  and  it  is  readily  proved 
that  two  tetraedrons  similarly  placed  in  the  two  polyedrons  are 
similar. 

We  leave  the  details  of  the  proof  to  the  student.     See  (III.  39). 

69.  Corollary.  Homologous  diagonals,  and  in  general  any  two  homol 
ogous  lines,  in  two  similar  polyedrons,  are  in  the  same  ratio  as  any  two 
homologous  edges,  that  is,  in  the  ratio  of  similitude  of  the  polyedrons. 


PKOPOSITION  XXVI.— THEOEEM. 

70.   Two  polyedrons  composed  of  the  same  number  of  tetraedrons, 
similar  each  to  each  and  similarly  placed,  are  similar. 
The  proof  is  left  to  the  student.     See  (III.  38). 


224 


GEOMETRY. 


PROPOSITION  XXVII.— THEOREM. 

71.  Similar  polyedrons  are  to  each  other  as  the  cubes  of  their  homol 
ogous  edges. 

1st.  Let  ABCD,  abed,   be  two  A  a 

similar  tetraedrons ;  let  the  similar 
faces  BCD,  bed,  be  taken  as  bases, 
and  let  A  0,  ao  be  their  altitudes. 

Since  the  tetraedrons  are  simi 
lar,  they  may  be  placed  with  their 
equal  homologous  polyedral  angles 
A  and  a  in  coincidence,  and  the 
base  bed  will  then  be  parallel  to 
the  base  BCD,  since  their  planes 

make  equal  angles  with  the  plane  of  the  face  ABC.  The  perpen 
dicular  AO,  to  BCD,  will  also  be  perpendicular  to  bed,  and  Ao  will 
be  the  altitude  of  the  tetraedron  Abed  or  abed.  Denoting  the 
volumes  of  the  tetraedrons  by  Fand  v,  we  have  (55), 

F_  BCD  X  AO  ^  BCD      AO 
v          bed  X  Ao  bed         Ao 

The  bases  being  similar,  we  have 

BCD       BC* 


bed          be2 


and  by  (69),  we  have 


AO  ^         ^ 

Ao         ae         be 


hence 


_  BC      BC\ 

v         be*    '  '    be          be 


or,  since  any  two  homologous  edges  are  in  the  same  ratio  as  any 
other  two,  the  two  similar  tetraedrons  are  to  each  other  as  the  cubes 
of  any  two  homologous  edges. 

2d.  Two  similar  polyedrons  may  be   decomposed  into  the  same 
number  of  tetraedrons,  similar  each  to  each ;  and  any  two  homologous 


BOOK    VII. 


225 


tetraedrons  are  to  each  other  as  the  cubes  of  their  homologous  edges  ; 
but  the  ratio  of  the  homologous  edges  of  the  two  similar  tetraedrons 
is  equal  to  ratio  of  any  two  homologous  edges  of  the  polyedron  (69) ; 
therefore,  any  two  homologous  tetraedrons  are  to  each  other  as  the 
cubes  of  two  homologous  edges  of  the  polyedron,  and  by  the  theory 
of  proportion,  their  sums,  or  the  polyedrons  themselves,  are  in  the 
same  ratio,  or  as  the  cubes  of  their  homologous  edges. 

72.  Corollary  I.  Similar  prisms,  or  pyramids  are  to  each  other  as 
the  cubes  of  their  altitudes. 

73.  Corollary  II.  Two  similar  polyedrons  are  to  each  other  as  the 
cubes  of  any  two  homologous  lines. 


SYMMETEICAL    POLYEDEONS. 


a.  Symmetry  with  respect  to  a  plane. 


I 


-r« 


74.  Definitions.  Two  points,  A  and  A',  are  sym 
metrical  with  respect  to  a  plane,  MN,  when  this 
plane  bisects  at  right  angles  the  straight  line  AA' 
joining  the  points. 

Two  figures  are  symmetrical  with  respect  to  a 
plane,  when  every  point  of  one  figure  has  its  sym-  JL 

-*4 

metrical  point  in  the  other. 

We  leave  the  proof  of  the   following  simple   theorems  to  the 
student. 


75.   Theorem.   The  symmetrical  figure  of  a  finite 
straight  line,  AB,  is  an  equal  straight  line,  A' B'. 


76.  Theorem.  The  symmetrical  figure  of 
an  indefinite  straight  line,  AB,  is  another 
indefinite  straight  line,  A  'B',  which  intersects 
the  first  in  the  plane  of  symmetry,  and 
makes  the  same  angle  with  the  plane. 


/f-r-f/ 


226 


GEOMETRY. 


77.  Theorem.   The  symmetrical  figure  of  a  plane  angle,  BA  C,  is  an 
equal  plane  angle,  B'A'C'  (Fig.  1). 

Fig.  3. 
Fig.  2.  B_ 


VIA 

E' 

78.  Theorem.  The  symmetrical  figure  of  a  plane  ABC,  is  a  plane 
ABC' ;  and  the  two  planes  intersect  in  the  plane  of  symmetry  ABN, 
and  make  equal  angles  with  it  (Fig.  2). 

Corollary.  If  a  plane  is  parallel  to  the  plane  of  symmetry,  its  sym 
metrical  plane  is  also  parallel  to  the  plane  of  symmetry,  and  at  the 
same  distance  from  it. 

79.  Theorem.    The  symmetrical  figure  of  a  diedral  angle,  CABD, 
is  an  equal  diedral  angle,  C'A'B'D'  (Fig.  3). 

PKOPOSITION  XXVIII.— THEOREM. 

80.  If  two  polyedrons  are  symmetrical  with  respect  to  a  plane,  1st, 
their  homologous  faces  are  equal;  2d,  their  homologous  polyedral  angles 
are  symmetrical. 

1st.  Let  A,  B,  C,  D,  be  the  vertices  of  a  face 
of  one  of  the  polyedrons ;  their  symmetrical 
points,  A',  B',  C',  D',  are  in  the  same  plane 
(78) ;  the  homologous  sides  of  the  polygons 
ABCD,  A'B'C'D',  are  equal  (75),  and  their 
homologous  angles  are  equal  (77)  ;  therefore 
the  homologous  faces  are  equal. 

2d.  The  homologous  face  angles  of  two 
polyedral  angles,  A  and  A ',  are  equal  (77), 
and  their  homologous  diedral  angles  are 
equal  (79)  ;  but  if  one  of  the  face  angles  as 


I 


BOOK   vii. 


227 


BAD  be  applied  to  its  equal  B'A'D',  so  as  to  bring  the  other  edges 
of  the  polyedral  angles  A  and  A'  on  the  same  side  of  the  common 
plane  B'A'D',  it  will  be  apparent  that  the  face  angles  succeed  each 
other  in  inverse  orders  in  the  two  figures  ;  therefore,  the  homologous 
polyedral  angles  of  the  two  polyedrons  are  symmetrical  (VI.  68). 

81.  Corollary.   Two  symmetrical  polyedrons  may  be  decomposed  into 
the  same  number  of  tetraedrons  symmetrical  each  to  each.     For  one  of 
the  polyedrons  being  divided  into  tetraedrons  by  drawing  diagonals 
from  a  common  vertex,  and  the  homologous  diagonals  being  drawn 
in  the   other   polyedron,  any  two  corresponding   tetraedrons    thus 
formed  will  have  their  vertices  symmetrical  each  to  each,  and  will 
consequently  be  symmetrical  tetraedrons. 

82.  Scholium.  Two  polyedrons  whose  faces  are  equal  each  to  each 
and  whose  polyedral  angles  are  symmetrical  each  to  each,  are  called 
symmetrical  polyedrons,  whatever  may  be  their  position  with  respect 
to  each  other,  since  they  admit  of  being  placed  on  opposite  sides  of  a 
plane  so  as  to  make  their  homologous  vertices   symmetrical  with 
respect  to  that  plane. 


PKOPOSITION  XXIX.—  THEOKEM. 

83.   Two  symmetrical  polyedrons  are  equivalent. 

Since  two  symmetrical  polyedrons  may  be  decomposed 
same  number  of  tetraedrons  symmetrical  each  to  each,  it 
necessary  to  prove  that  two  symmetrical  tetra 
edrons  are  equivalent. 

IjQtSABC  be  a  tetraedron;  let  the  plane 
of  one  of  its  faces,  ABC,  be  taken  as  a  plane 
of  symmetry,  and  construct  its  symmetrical 
tetraedron  8'  ABC.  The  tetraedrons,  having 
the  same  base  ABC  and  equal  altitudes  SO, 
S'O,  are  equivalent  (55). 


b.  Symmetry  with  respect  to  a  centre. 

84.  Definitions.  Two  points  A  and  A'  ,  are  sym- 
metrical  with  respect  to  a  fixed  point,  0,  called 
the  centre  of  symmetry,  when  this  point  bisects 
the  straight  line,  AA',  joining  the  two  points. 


into  the 
is  only 


.„••• 


228 


GEOMETRY. 


Any  two  figures  are  symmetrical  with  respect  to  a  centre,  when 
every  point  of  one  figure  has  its  symmetrical  point  on  the  other. 

These  definitions  are  identical  with  those  given  in  (I.  138),  but 
are  here  extended  to  figures  in  space. 

The  student  can  readily  establish  the  following  theorems  on  figures 
symmetrical  with  respect  to  a  centre. 

85.  Theorem.  The  symmetrical  figure  of  a  finite  straight  line,  AB, 
is  an  equal  straight  line,  A' B,'  parallel  to  the  first  (Fig.  1). 

Fig.  2. 
Fig.  1.  \B 


86.  Theorem.  The  symmetrical  figure  of  a  plane  angle,  BA  G,  is  an 
equal  plane  angle,  B'A'C'  (Fig.  2). 

87.  Theorem.  The  symmetrical  figure  of  a  plane,  BAG,  is  a  parallel 
plane,  B' A' G'  (Fig.  2). 


88.  Theorem.  The  symmetrical 
figure  of  a  diedral  angle,  DAB  G,  is 
an  equal  diedral  angle,  D'A'B' Gr. 


89.  Theorem.  If  two  polye- 
drons  are  symmetrical  with  re 
spect  to  a  centre,  1st,  their  ho 
mologous  faces  are  equal;  2d, 
their  homologous  angles  are  sym 
metrical. 

Corollary  I.  The  symmetrical  figure  of  a  polyedron  is  the  same, 
whether  the  symmetry  be  with  respect  to  a  plane  or  with  respect  to  a 
centre. 

Corollary  II.  Two  polyedrons,  symmetrical  with  respect  to  a  centre, 
are  equivalent. 


BOOK    VII. 


229 


c.  Symmetry  of  a  single  figure. 

90.  Definition.  Any  figure  in  space  is  called  a,  symmetrical  figure, 
1st,  if  it  can  be  divided  by  a  plane  into  two  figures  which  are  sym 
metrical  with  respect  to  that  plane;  2d,  if  it  has  a  centre  which 
bisects  all  straight  lines  drawn  through  it,  and  terminated  by  the  sur 
face  of  the  figure ;  3d,  if  it  has  an  axis  which  contains  the  centres 
of  all  the  sections  perpendicular  to  that  axis. 

For  example,  1st,  the  hexaedron  SABCS' 
is  symmetrical  with  respect  to  the  plane  ABC, 
which  divides  the  solid  into  the  two  symmet 
rical  tetraedrons  SAB  C,  S'AB  C. 

2d.  The  intersection  of  the  four  diagonals 
of  a  parallelepiped  is  the  centre  of  symmetry 
of  the  parallelepiped  (18). 

3d.  The  straight  line  zz't  joining  the  cen 
tres  of  the  bases  of  a  right  parallelepiped 
AC',  is  an  axis  of  symmetry  of  the  figure, 
since  it  evidently  contains  the  centre  0  of  any 
section  abed  perpendicular  to  it,  or  parallel  to 
the  bases.  If  the  parallelepiped  is  rectangu 
lar,  it  has  three  axes  xx',  yy',  zz',  perpendicu 
lar  to  each  other  which  intersect  in  its  centre. 

We  leave  the  demonstration  of  the  following  theorems  to  the 
student. 


91.  Theorem.  If  a  figure  has  two 
planes  of  symmetry,  MN  and  PQ, 
the  intersection,  xx',  of  these  planes, 
is  an  axis  of  symmetry  of  the  figure. 

See  (I.  141). 


20 


230 


GEOMETRY. 


92.  Theorem.  If  a  figure  has  three  planes  of  symmetry  perpendicular 
to  each  other  (VI.  48),  the  intersections  of  these  planes  are  three  axes 
of  symmetry,  and  the  common  intersection  of  these  axes  is  the  centre  of 
symmetry  of  the  figure. 


THE    EEGULAE   POLYEDEONS. 

93.  Definition.  A  regular  polyedron  is  one  whose  faces  are  all  equal 
regular  polygons  and  whose  polyedral  angles  are  all  equal  to  each 
other. 

PROPOSITION  XXX.— PEOBLEM. 

94.  To  construct  a  regular  polyedron,  having  given  one  of  its  edges. 
There  are  five  regular  polyedrons,  which  we  shall  consider  in  their 

order. 

Construction  of  the  regular  tetraedron. 

Let  AB  be  the  given  edge.  Upon  AB  con 
struct  the  equilateral  triangle  ABC.  At  the 
centre  0  of  this  triangle  erect  a  perpendicular, 
OD,  to  its  plane,  and  take  the  point  D  so  that 
AD  =  AB-,  join  DA,  DB,  DC.  The  faces  of 
the  tetraedron  ABCD  are  each  equal  to  the  face 
ABC  (VI.  10),  and  its  polyedral  angles  are  all 
equal  (VI.  71);  therefore,  AB  CD  is  a  regular 
tetraedron.  ^ 

Construction  of  the  regular  hexaedron, 

Upon  the  given  edge  AB,  construct  the  square 
ABCD.  The  cube  ABCDE,  whose  faces  are  each 
equal  to  this  square,  is  a  regular  hexaedron,  and  the 
method  of  constructing  it  is  obvious. 

Construction  of  the  regular  octaedron. 

Let  AB  be  the  given  edge.  Upon  AB  construct  the  square 
ABCD,  and  at  the  centre  0  of  the  square  erect  the  perpendicular 
FO  to  its  plane.  In  this  perpendicular,  take  the  points  F  and  G  so 


BOOK     VII. 


231 


that  OF=  OA  and  OG  =  OA,  and  join  FA, 
FB,  FC,  FD,  GA,  GB,  GC,  GD.  These 
edges  are  equal  to  each  other  (VI.  10),  and 
also  to  the  edge  AB,  since  A  OF  and  A  OB 
are  equal  triangles ;  therefore,  the  faces  of  the 
figure  are  eight  equal  equilateral  triangles. 

Since  the  triangles  DFB  and  DAB  are 
equal,  DFBG  is  a  square,  and  it  is  evident 
that  the  pyramid  A-DFBG  is  equal  in  all  its  parts  to  the  pyramid 
F-ABCD]  therefore,  the  polyedral  angles  A  and  F  are  equal; 
whence,  also,  all  the  polyedral  angles  of  the  figure  are  equal  to  each 
other,  and  the  figure  is  a  regular  octaedron. 


Construction  of  the  regular  dodecaedron. 

Upon  the  given  edge  AB,  construct  a  regular  pentagon  ABODE-, 
to  each  of  the  sides  of  this  pentagon  apply  the  side  of  an  equal 

F 


pentagon,  and  let  the  planes  of  these  pentagons  be  so  inclined  to 
that  of  ABODE  as  to  form  triedral  angles  at  A,  B,  0,  D,  E.  There 
is  thus  formed  a  convex  surface,  FGHI,  etc.,  composed  of  six  regu 
lar  pentagons. 

Construct  a  second  convex  surface,  F'G'H'I',  etc.,  equal  to  the 
first.  The  two  surfaces  may  be  combined  so  as  to  form  a  single  con 
vex  surface.  For,  suppose  the  diagram  to  represent  the  exterior  of 
the  first  surface  and  the  interior  of  the  second ;  let  the  point  P  of 
the  first  be  placed  on  F'  of  the  second ;  then  the  three  equal  angles 
OFF,  P'F'A',  A'F'G',  can  be  united  so  as  to  form  a  triedral  angle 
at  F'  equal  to  that  at  A',  since  the  diedral  angle  F'A'  is  already 
that  which  belongs  to  such  a  triedral  angle.  But  when  PF  coin 
cides  with  F'G',  there  will  be  brought  together  at  G'  three  angles 
PFA,  AFG,  F'G'H1,  which  will  form  a  triedral  angle  equal  to  A', 


232  GEOMETRY. 

since  the  diedral  angles  at  the  edges  FA  and  F'  G '  are  already  those 
which  belong  to  such  an  angle.  Thus,  it  can  be  shown,  successively, 
that  all  the  edges  PF,  FG,  etc.,  of  the  first  figure,  will  coincide  with 
the  edges  F'G',  G'H',  etc.,  of  the  second,  and  that  all  the  polyedral 
angles  of  the  whole  convex  surface  thus  formed  are  equal.  This 
surface  is  therefore  a  regular  dodecaedron. 

Construction  of  the  regular  icosaedron. 

Upon  the  given  edge  AB,  construct  a  regular  pentagon  ABODE, 
and  at  its  centre  0  erect  OS  perpendicular  to  its  plane,  taking  S  so 
that  SA  =  AB ;  then,  joining  SA,  SB,  etc.,  the  pyramid  S- ABODE 
is  regular,  and  each  of  its  faces  is  an  equilateral  triangle.  Now  let 


the  vertices  A  and  B  be  taken  (as  in  the  second  figure)  as  the  vertices 
of  two  other  pyramids,  A-BSEFG  and  B-ASCHG,  each  equal  to 
the  first  and  having  in  common  with  it  the  faces  ASB  and  ASE, 
ASB  and  BSC,  respectively,  and  in  common  with  each  other  the 
faces  ASB  and  ABG.  There  is  thus  formed  a  convex  surface 
CDEFGH,  composed  of  ten  equal  equilateral  triangles. 

Construct  a  second  convex  surface  C'D'E'F' G'H', equal  in  all  re 
spects  to  the  first ;  and  let  the  figures  represent  the  exterior  of  the  first 
surface,  and  the  interior  of  the  second.  Let  the  first  surface  be  applied 
to  the  second  by  bringing  the  point  D,  where  two  faces  meet,  upon  the 
point  0',  where  three  faces  meet.  The  edges  DE  and  D(7can  then 
be  brought  into  coincidence  with  the  edges  C'D'  and  C'H',  re 
spectively,  to  form  a  polyedral  angle  of  five  faces  equal  to  8,  without 
in  any  way  changing  the  form  of  either  surface,  since  the  diedral 
angles  at  the  edges  SD,  S'C',B'Cf,  are  those  which  belong  to  such 
a  polyedral  angle.  But  when  D(7has  been  brought  into  coincidence 
with  C'H',  there  have  been  brought  together,  at  the  point  H',  five 


BOOK   vii.  233 

equal  faces  having  the  necessary  diedral  inclinations  to  form  another 
polyedral  angle  equal  to  $;  and  thus,  in  succession,  it  can  be  shown 
that  all  the  outer  edges  of  the  first  surface  coincide  with  those  of  the 
second,  and  that  all  the  polyedral  angles  of  the  entire  convex  sur 
face  thus  formed  are  equal.  This  surface  is  therefore  a  regular 
icosaedron. 


PEOPOSITION  XXXI.— THEOEEM. 

95.   Only  five  regular  (convex)  polyedrons  are  possible. 

The  faces  of  a  regular  polyedron  must  be  regular  polygons,  and 
at  least  three  faces  are  necessary  to  form  a  polyedral  angle. 

1st.  The  simplest  regular  polygon  is  the  equilateral  triangle. 
Three  angles  of  an  equilateral  triangle  can  be  combined  to  form  a 
convex  polyedral  angle,  and  this  combination,  as  shown  in  the  pre 
ceding  proposition,  gives  the  regular  tetraedron. 

The  combination  of  four  such  angles  gives  the  regular  octaedron ; 
and  that  of  five  gives  the  regular  icosaedron.  The  combination  of 
six  or  more  (each  being  -J  of  a  right  angle)  gives  a  sum  equal  to,  or 
greater  than,  four  right  angles,  and  therefore  cannot  form  a  convex 
polyedral  angle  (VI.  70).  Therefore,  only  three  regular  convex 
polyedrons  are  possible  whose  surfaces  are  composed  of  triangles. 

2d.  Three  right  angles  can  be  combined  to  form  a  polyedral  angle, 
and  this  combination  gives  the  regular  hexaedron,  or  cube.  Four 
or  more  right  angles  cannot  form  a  convex  polyedral  angle  (VI.  70) ; 
therefore,  but  one  regular  convex  polyedron  is  possible  whose  surface 
is  composed  of  squares. 

3d.  Three  angles  of  a  regular  pentagon,  being  less  than  four  right 
angles  (each  being  f  of  a  right  angle),  may  form  a  polyedral  angle, 
as  in  the  case  of  the  dodecaedron ;  but  four  or  more  would  exceed 
four  right  angles.  Therefore,  but  one  regular  convex  polyedron  is 
possible  with  pentagonal  faces. 

4th.  Three  or  more  angles  of  a  regular  hexagon  (each  being  f  of 
a  right  angle)  cannot  form  a  convex  polyedral  angle ;  nor  can  angles 
of  any  regular  polygon  of  a  greater  number  of  sides  form  such  a 
polyedral  angle. 

Therefore,  the  five  regular  convex  polyedrons  constructed  in  the 
preceding  proposition  are  the  only  ones  possible. 

20* 


234 


GEOMETRY. 


96.  Scholium.  The  student  may  derive  some  aid  in  comprehending 
the  preceding  discussion  of  the  regular  polyedrons  by  constructing 
models  of  them,  which  he  can  do  in  a  very  simple  manner,  and  at 
the  same  time  with  great  accuracy,  as  follows. 

Draw  on  card-board  the  following  diagrams ;  cut  them  out  entire, 
and  at  the  lines  separating  adjacent  polygons  cut  the  card-board 
half  through ;  the  figures  will  then  readily  bend  into  the  form  of  the 
respective  surfaces,  and  can  be  retained  in  that  form  by  glueing  the 


llexaedron. 


Octaedron. 


Tetraedron. 


Dodeoaedron. 


Icosaedron. 


GENERAL   THEOREMS    ON    POLYEDRONS. 
PROPOSITION  XXXII.— THEOREM. 

97.   In  any  polyedron,  the  number  of  its  edges  increased  by  two 
is  equal  to  the  number  of  its  vertices  increased  by  the  number  of  its 

faces. 


BOOK     VII. 


235 


Let  E  denote  the  number  of  edges  of  any  polyedron,  V  the  num 
ber  of  its  vertices,  and  F  the  number  of  its  faces  ;  then  we  are  to 
prove  that 


In  the  first  place,  we  observe  that  if  we 
remove  a  face,  as  ABODE,  from  any  con 
vex  polyedron  GH,  we  leave  an  open  sur 
face,  terminated  by  a  broken  line  which 
was  the  contour  of  the  face  removed  ;  and 
in  this  open  surface  the  number  of  edges 
and  the  number  of  vertices  remain  the 
same  as  in  the  original  surface. 

Now  let  us  form  this  open  surface  by  putting  together  its  faces 
successively,  and  let  us  examine  the  law  of  connection  between  the 
number  of  edges  E,  the  number  of  vertices  F,  and  the  number  of 
faces,  at  each  successive  step.  Beginning  with  one  face  we  have 
E=V.  Annexing  a  second  face,  by  applying  one  of  its  edges  to  an 
edge  of  the  first,  we  form  a  surface  having  one  edge  and  two  vertices 
in  common  with  the  first  ;  therefore,  whatever  the  number  of  sides 
of  the  new  face,  the  whole  number  of  edges  is  now  one  more  than 
the  whole  number  of  vertices  ;  that  is, 


For  2  faces, 


E=  7+1. 


Annexing  a  third  face,  adjacent  to  each  of  the  former,  the  new  sur 
face  will  have  two  edges  and  three  vertices  in  common  with  the  pre 
ceding  surface;  therefore  the  increase  in  the  number  of  edges  is 
again  one  more  than  the  increase  in  the  number  of  vertices ;  and  we 
have 

For  3  faces,  E=V+2. 

At  different  stages  of  this  process  the  number  of  common  edges  to 
two  successive  open  surfaces  may  vary,  but  in  all  cases  it  is  ap 
parent  that  the  addition  of  a  new  face  increases  E  by  one  more  unit 
than  it  increases  F;  and  hence  we  have  the  following  series  of 
results : 


236  GEOMETRY. 


In  an  open  surface  of 


1  face,  E  =  7, 

2  faces,  E  =  V  +  1, 

3  «  #=74-2, 

4  «  .£==  74-3, 
etc.  etc. 

.F— 1  faces,  E=V+F—2; 


where  the  law  is,  that,  in  the  successive  values  of  E,  the  number  to 
be  added  to  7  is  a  unit  less  than  the  number  of  faces.  The  last  line 
expresses  the  relation  for  the  open  surface  of  F  —  1  faces,  that  is, 
for  the  open  surface  which  wants  but  one  face  to  make  the  closed  sur 
face  of  F  faces.  But  the  number  of  edges  and  the  number  of  ver 
tices  of  this  open  surface  are  the  same  as  in  the  closed  surface. 
Therefore,  in  a  closed  surface  of  F  faces,  we  have 

E=  7+.F— 2, 
or 

E+2  =  V+F, 

as  was  to  be  proved. 

This  theorem  was  discovered  by  Euler,  and  is  called  Eider's  Theo 
rem  on  Polyedrons. 


PEOPOSITION  XXXIII.— THEOEEM. 

98.  The  sum  of  all  the  angles  of  the  faces  of  any  polyedron  is  equal 
to  four  right  angles  taken  as  many  times  as  the  polyedron  has  vertices 
less  two. 

Let  E  denote  the  number  of  edges,  7  the  number  of  vertices,  F 
the  number  of  faces,  and  S  the  sum  of  all  the  angles  of  the  faces,  of 
any  polyedron. 

If  we  consider  both  the  interior  angles  of  a  polygon  and  the 
exterior  ones  formed  by  producing  its  sides  as  in  (I.  101),  the  sum  of 
all  the  angles  both  interior  and  exterior  is  2R  X  n,  where  R  denotes 
a  right  angle,  and  n  is  the  number  of  sides  of  the  polygon.  If, 
then,  E  denotes  the  number  of  edges  of  the  polyedron,  2E  denotes 
the  whole  number  of  sides  of  all  its  faces  considered  as  independent 
polygons,  and  the  sum  S  of  the  interior  angles  of  all  the  F  faces 
plus  the  sum  of  their  exterior  angles  is  2R  X  2J£.  But  the  sum  of 


BOOK  vn.  237 

the  exterior  angles  of  one  polygon  is  4jR,  and  the  sum  of  the  exterior 
angles  of  the  F  polygons  is  4R  X  F\  that  is, 

S  +  4R  X  F  =  2R  X  2E, 
or,  reducing, 

fl=4RX  (E—F). 

But  by  Euler's  Theorem,  E  —  F  =  V—  2;  Hence, 
S=±KX  (F— 2). 


BOOK   VIII. 

THE  THREE  ROUND  BODIES. 

OF  the  various  solids  bounded  by  curved  surfaces,  but  three  are 
treated  of  in  Elementary  Geometry — namely,  the  cylinder,  the  cone, 
and  the  sphere,  which  are  called  the  THREE  ROUND  BODIES. 

THE  CYLINDER. 

2.  Definition.  A  cylindrical  surface  is  a  curved  surface  generated 
by  a  moving  straight  line  which  continually  touches  a  given  curve, 
and  in  all  of  its  positions  is  parallel  to  a  given  fixed  straight  line  not 
in  the  plane  of  the  curve. 

Thus,  if  the  straight  line  Aa  moves  so 
as  continually  to  touch  the  given  curve 
ABCD,  and  so  that  in  any  of  its  positions, 
as  Bb,  Cc,  Dd,  etc.,  it  is  parallel  to  a 
given  fixed  straight  line  Mm,  the  surface 
ABCDdcba  is  a  cylindrical  surface.  If 
the  moving  line  is  of  indefinite  length,  a 
surface  of  indefinite  extent  is  generated. 

The  moving  line  is  called  the  generatrix ;  the  curve  which  it  touches 
is  called  the  directrix.  Any  straight  line  in  the  surface,  as  Bb,  which 
represents  one  of  the  positions  of  the  generatrix,  is  called  an  element 
of  the  surface. 

In  this  general  definition  of  a  cylindrical  surface,  the  directrix 
may  be  any  curve  whatever.  Hereafter  we  shall  assume  it  to  be  a 
closed  curve,  and  usually  a  circle,  as  this  is  the  only  curve  whose 
properties  are  treated  of  in  elementary  geometry. 

238 


BOOK     VIII. 


239 


3.  Definition.  The  solid  Ad  bounded  by  a  cylindrical  surface  and 
two  parallel  planes,  ABD  and  abd,  is  called  a  cylinder;  its  plane 
surfaces,  ABD,  abd,  are  called  its  bases ;  the  curved  surface  is  some 
times  called  its  lateral  surface;  and  the  perpendicular  distance  be 
tween  its  bases  is  its  altitude. 

A  cylinder  whose  base  is  a  circle  is  called  a  circular  cylinder. 

4.  Definition.  A  right  cylinder  is  one  whose  ele 
ments  are  perpendicular  to  its  base. 

5.  Definition.   A   right   cylinder  with   a   circular 
base,  as  ABCa,  is  called  a  cylinder  of  revolution,  be 
cause  it  may  be  generated  by  the  revolution  of  a 
rectangle  A  Ooa  about  one  of  its  sides,  Oo,  as  an 
axis;  the  side  Aa  generating  the  curved  surface, 

and  the  sides  OA  and  oa  generating  the  bases.  The  fixed  side 
Oo  is  the  axis  of  the  cylinder.  The  radius  of  the  base  is  called  the 
radius  of  the 


PEOPOSITION  I.— THEOEEM. 

6.  Every  section  of  a  cylinder  made  by  a  plane  passing  through  an 
element  is  a  parallelogram. 

Let  Bb  be  an  element  of  the  cylinder  Ac ; 
then,  the  section  BbdD,  made  by  a  plane 
passed  through  Bb,  is  a  parallelogram. 

1st.  The  line  Dd  in  which  the  cutting  plane 
intersects  the  curved  surface  a  second  time  is 
an  element.     For,  if  through  any  point  D  of 
this  intersection  a  straight  line  is  drawn  paral 
lel  to  Bb,  this  line  by  the  definition  of  a  cylindrical  surface,  is  an 
element  of  the  surface,  and  it  must  also  lie  in  the  plane  Bd;  there 
fore,  this  element,  being  common  to  both  surfaces,  is  their  inter 
section. 

2d.  The  lines  BD  and  bd  are  parallel  (VI.  25),  and  the  elements 
Bb  and  Dd  are  parallel ;  therefore,  Bd  is  a  parallelogram. 

7.  Corollary.  Every  section  of  a  right  cylinder  made  by  a  plane 
perpendicular  to  its  base  is  a  rectangle. 


240 


GEOMETRY. 


PROPOSITION  II.— THEOREM. 

8.   The  bases  of  a  cylinder  are  equal. 

Let  BD  be  the  straight  line  joining  any 
two  points  of  the  perimeter  of  the  lower  base, 
and  let  a  plane  passing  through  BD  and  the 
element  Bb  cut  the  upper  base  in  the  line  bd ; 
then,  BD  ==  bd  (6). 

Let  A  be  any  third  point  in  the  perimeter 
of  the  lower  base,  and  Aa  the  corresponding 

element.  Join  AB,  AD,  ab,  ad.  Then  AB  —  ab  and  AD  =  ad 
(6)  ;  and  the  triangles  ABD,  abd,  are  equal.  Therefore,  if  the  upper 
base  be  applied  to  the  lower  base  with  the  line  bd  in  coincidence 
with  its  equal  BD,  the  triangles  will  coincide  and  the  point  a  will 
fall  upon  A ;  that  is,  any  point  a  of  the  upper  base  will  fall  on  the 
perimeter  of  the  lower  base,  and  consequently  the  perimeters  will 
coincide  throughout.  Therefore,  the  bases  are  equal. 


9.  Corottary.I.  Any  two  parallel  sections 
MPN,  mpn,  of  a  cylindrical  surface  Ab,  are 
equal. 

For,  these  sections  are  the  bases  of  the 
cylinder  Mn. 


10.  Corollary  II.  All  the  sections  of  a  circular  cylinder  parallel 
to  its  bases  are  equal  circles ;  and  the  straight  line  joining  the  centres 
of  the  bases  passes  through  the  centres  of  all  the  parallel  sections. 
This  line  is  called  the  axis  of  the  cylinder. 

11.  Definition.  A  tangent  plane  to  a  cylinder  is  a  plane  which 
passes  through  an  element  of  the  curved  surface  without  cutting  this 
surface.     The  element  through  which  it  passes  is  called  the  element 
of  contact. 


BOOK    VIII.  241 


PEOPOSITION  III.— PKOBLEM. 

12.  Through  a  given  point,  to  pass  a  plane  tangent  to  a  given  circular 
cylinder. 

1st.  When  the  given  point  is  in  the  curved  surface  of  the  cylinder, 
in  which  case  the  element  of 
contact  is  given,  since  it  must 
be  the  element  passing  through 
the  given  point. 

Let  the  given  point  be  a 
point  in  the  element  Aa.  At 
A,  in  the  plane  of  the  base, 
draw  A  T  tangent  to  the  base, 
and  pass  a  plane  Rt  through 
Aa  and  A  T;  this  plane  is  tan 
gent  to  the  cylinder.  For,  let  P  be  any  point  in  this  plane  not  in 
the  element  Aa,  and  through  P  pass  a  plane  parallel  to  the  base,  in 
tersecting  the  cylinder  in  the  circle  MN  and  the  plane  Rt  in  the  line 
MP.  Let  Q  be  the  centre  of  the  circle  MN,  and  join  QM.  Since 
MPand  MQ  are  parallel  respectively  to  AT  and  AO  (VI.  25),  the 
angle  PMQ  is  equal  to  the  angle  TAO,  and  PM  is  tangent  to  the 
circle  MN  at  M ;  therefore,  P  lies  without  the  circle  MN  and  conse 
quently  without  the  cylinder.  Hence  the  plane  Rt  does  not  cut  the 
cylinder  and  is  a  tangent  plane. 

2d.  When  the  given  point  is  without  the  cylinder.  Let  P  be  the 
given  point.  Through  P  draw  the  straight  line  PT,  parallel  to  the 
elements  of  the  cylinder,  meeting  the  plane  of  the  base  in  T.  From 
T  draw  TA  and  TC  tangents  to  the  base  (II.  90) ;  through  PT  and 
the  tangent  TA  pass  a  plane  Rt,  and  through  PT  and  TC  pass  a 
plane  Ts.  The  plane  Rt,  passing  through  PTand  the  point  A,  must 
contain  the  element  Aa,  since  Aa  is  parallel  to  PT;  and  it  is  a  tan 
gent  plane  since  it  also  contains  the  tangent  A  T.  For  a  like  reason 
the  plane  Ts  is  a  tangent  plane. 

13.  Corollary.  The  intersection  of  two  tangent  planes  to  a  cylinder 
is  parallel  to  the  elements  of  the  cylinder. 

14.  /Scholium.  Any  straight  line,  drawn  in  a  tangent  plane  and 
cutting  the  element  of  contact,  is  tangent  to  the  cylinder. 

21  Q 


242 


GEOMETRY. 


THE    CONE. 

15.  Definition.  A  conical  surface   is   a   curved  surface   generated 
by  a  moving  straight  line  which  continually  touches  a  given  curve, 
and  passes   through  a  given  fixed  point  not  in  the  plane  of  the 
curve. 

Thus,  if  the  straight  line  SA  moves  so 
as  continually  to  touch  the  given  curve 
AS  CD,  and  in  all  its  positions,  SB,  SC, 
SD,  etc.,  passes  through  the  given  fixed 
point  S,  the  surface  S-AB  CD  is  a  conical 
surface. 

The  moving  line  is  called  the  generatrix; 
the  curve  which  it  touches  is  called  the 
directrix.  Any  straight  line  in  the  surface, 

as  SB,  which  represents  one  of  the  positions  of  the  generatrix,  is 
called  an  element  of  the  surface.     The  point  S  is  called  the  vertex. 

If  the  generatrix  is  of  indefinite  length,  as  AS  a,  the  whole  surface 
generated  consists  of  two  symmetrical  portions,  each  of  indefinite 
extent,  lying  on  opposite  sides  of  the  vertex,  as  S-AB  CD  and 
S-abcd,  which  are  called  nappes;  one  the  upper,  the  other  the  lower 
nappe. 

16.  Definition.  The  solid  S-ABCD,  bounded  by  a  conical  surface 
and  a  plane  ABD  cutting  the  surface,  is  called  a  cone;  its  plane  sur 
face  ABD  is  its  base,  the  point  8  is  its  vertex,  and  the  perpendicular 
distance  SO  from  the  vertex  to  the  base  is  its  altitude. 

A  cone  whose  base  is  a  circle  is  called  a  circular  done.  The  straight 
line  drawn  from  the  vertex  of  a  circular  cone  to  the  centre  of  its 
base  is  the  axis  of  the  cone. 

17.  Definition.  A  right  circular  cone  is  a  circular 
cone   whose   axis   is   perpendicular    to   its   base,    as 
S-ABCD. 

The  right  circular  cone  is  also  called  a  cone  of  revo 
lution,  because  it  may  be  generated  by  the  revolution 
of  a  triangle,  SAO,  about  one  of  its  perpendicular 
sides,  SO,  as  an  axis;  the  hypotenuse  SA  gener 
ating  the  curved  surface,  and  the  remaining  perpen 
dicular  side  OA  generating  the  base. 


BOOK     VIII. 


243 


PROPOSITION  IV.— THEOEEM. 

18.  Every  section  of  a  cone  made  by  a  plane  passing  through  its  ver 
tex  is  a  triangle. 

Let  the  cone  S-ABCD  be  cut  by  a  plane  SBC  which  passes 
through  the  vertex  $  and  cuts  the  base  in  the  straight  line  BC\ 
then,  the  section  SBC  is  a  triangle,  that  is,  the 
intersections  SB  and  SC  with  the  curved  surface 
are  straight  lines. 

For,  the  straight  lines  joining  8  with  B  and  C 
are  elements  of  the  surface,  by  the  definition  of  a 
cone,  and  they  also  lie  in  the  cutting  plane; 
therefore  they  coincide  with  the  intersections  of 
that  plane  with  the  curved  surface. 


PROPOSITION  V.—  THEOREM. 

19.  If  the  base  of  a  cone  is  a  circle,  every  section  made  by  a  plane 
parallel  to  the  base  is  a  circle. 

Let  the  section  abc,  of  the  circular  cone 
S-ABC,  be  parallel  to  the  base. 

Let  0  be  the  centre  of  the  base,  and  let  o 
be  the  point  in  which  the  axis  SO  cuts  the 
plane  of  the  parallel  section.  Through  SO 
and  any  number  of  elements  SA,  SB,  etc., 
pass  planes  cutting  the  base  in  the  radii  OA, 
OB,  etc.,  and  the  parallel  section  in  the 
straight  lines  oa,  ob,  etc.  Since  oa  is  parallel  to  OA,  and  ob  to  OB, 
we  have 


oa 


So 


ob 


So 


oa 


ob 


But  OA  =  OB,  therefore  oa  —  ob  ;  hence,  all  the  straight  lines 
drawn  from  o  to  the  perimeter  of  the  section  are  equal,  and  the  sec 
tion  is  a  circle. 

20.  Corollary.   The  axis  of  a  circular  cone  passes  through  the 
centres  of  all  the  sections  parallel  to  the  base. 


244  GEOMETRY. 

21.  Definition.  A  tangent  plane  to  a  cone  is  a  plane  which  passes 
through  an  element  of  the  curved  surface  without  cutting  this  sur 
face.     The  element  through  which  it  passes  is  called  the  element  of 
contact. 

PROPOSITION  VI—  PROBLEM. 

22.  Through  a  given  point,  to  pass  a  plane  tangent  to  a  given  circular 
cone. 

1st.  When  the  given  point  is  in  the  curved  surface  of  the  cone. 


Let  the  given  point  be  a  point  in  the  element  SA.  At  A,  in  the 
plane  of  the  base,  draw  AM  tangent  to  the  base,  and  pass  a  plane 
MP  through  SA  and  AM ;  this  plane  is  tangent  to  the  cone.  The 
proof  is  the  same  as  for  the  tangent  plane  to  the  cylinder. 

2d.  When  the  given  point  is  a  point  m  without  the  cone.  Join 
the  vertex  S  and  the  point  m,  and  produce  Sm  to  meet  the  plane  of 
the  base  in  M.  From  M  draw  MA  and  M C,  tangents  to  the  base, 
and  through  SM  and  these  tangents  pass  the  planes  MP  and  ME. 
The  plane  MP,  containing  the  element  SA  and  the  tangent  MA,  is  a 
tangent  plane  to  the  cone,  and  it  also  passes  through  the  given 
point  m  ;  and  for  a  like  reason,  the  plane  ME  also  satisfies  the  con 
ditions  of  the  problem. 

23.  Scholium  I.  Any  straight  line,  drawn  in  a  tangent  plane  and 
cutting  the  element  of  contact,  is  tangent  to  the  cone. 

24.  Scholium  II.  When  the  given  point  is  without  the  cone,  the 
problem  may  be  stated  in  the  following  form : 

Through  any  given  straight  line  passing  through  the  vertex  of  a  cone, 
to  pass  a  plane  tangent  to  the  cone. 


BOOK     VIII. 


245 


THE  SPHEEE. 

25.  Definition.  A  sphere  is  a  solid  bounded  by  a  surface  all  the 
points  of  which  are  equally  distant  from  a  point  within  called  the 
centre. 

A  sphere  may  be  generated  by  the  revolution 
of  a  semicircle  ABC  about  its  diameter  A  C  as  an 
axis  ;  for  the  surface  generated  by  the  curve  ABC 
will  have  all  its  points  equally  distant  from  the 
centre  0. 

A   radius  of  the  sphere  is   any  straight   line 
drawn  from  the  centre  to  the  surface.    A  diameter 
is  any  straight  line  drawn  through  the  centre  and  terminated  both 
ways  by  the  surface. 

Since  all  the  radii  are  equal  and  every  diameter  is  double  the 
radius,  all  the  diameters  are  equal. 

26.  Definition.  It  will  be  shown  that  every  section  of  a  sphere 
made  by  a  plane  is  a  circle ;  and  as  the  greatest  possible  section  is 
one  made  by  a  plane  passing  through  the  centre,  such  a  section  is 
called  a  great  circle.     Any  section  made  by  a  plane  which  does  not 
pass  through  the  centre  is  called  a  small  circle. 

27.  Definition.  The  poles  of  a  circle  of  the  sphere  are  the  extremi 
ties  of  the  diameter  of  the  sphere  which  is  perpendicular  to  the  plane 
of  the  circle ;  and  this  diameter  is  called  the  axis  of  the  circle. 


PEOPOSITION  VIL— THEOEEM. 

28.  Every  section  of  a  sphere  made  by  a  plane  is  a  circle 
Let  abc  be  a  plane  section  of  the  sphere 

whose  centre  is  0. 

All  the  straight  lines  Oa,  Ob,  etc.,  drawn 
from  0  to  points  in  the  curve  of  intersec 
tion  abc,  are  equal,  being  radii  of  the 
sphere ;  therefore,  the  curve  abc  is  the  cir 
cumference  of  a  circle  (VI.  12),  and  its 
centre  is  the  foot  o  of  the  perpendicular  Oo 
let  fall  from  0  upon  the  plane  of  the  section. 

29.  Corollary  I.  All  great  circles,  as  ABC,  ADCE,  are  equal; 


246 


GEOMETRY. 


for,  since  their  planes  pass  through  the  centre  of  the  sphere,  their 
radii  OA,  Oa,  are  radii  of  the  sphere. 

30.  Corollary  II.   A  small   circle  abe  is  the  less,  the  greater  its 
distance  Oo  from  the  centre  of  the  sphere. 

31.  Corollary  III.  Every  great  circle  divides  the  sphere  into  two 
equal  parts ;  for,  if  the  parts  be  separated  and  then  placed  with  their 
bases  in  coincidence  and  their  convexities  turned  the  same  way,  their 
surfaces  will  coincide ;  otherwise  there  would  be  points  in  the  spheri 
cal  surface  unequally  distant  from  its  centre. 

32.  Corollary  IV.  Any  two  great  circles  A  CBD.  AEBF,  bisect  each 
other;  for,  the  common  intersection  AB  of 

their  planes  passes  through  the  centre  of  the 
sphere  and  is  a  diameter  of  each  circle. 

33.  Corollary  V.  An  arc  of  a  great  circle  may 
be  drawn  through  any  two  given  points,  A,  E, 
of  the  surface  of  the  sphere ;  for  the  two  points, 
A  and  E,  together  with  the  centre  0,  deter 
mine  the  plane  of  a  great  circle  whose  cir 
cumference  passes  through  A  and  E  (VI.  4). 

If,  however,  the  two  given  points  are  the  extremities  A  and  B  of 
a  diameter  of  the  sphere,  the  position  of  the  circle  is  not  determined, 
for  the  points  A,  0  and  B,  being  in  the  same  straight  line,  an  infi 
nite  number  of  planes  can  be  passed  through  them  (VI.  2). 

34.  Corollary  VI.  An  arc  of  a  circle  may  be  drawn  through  any 
three  given  points  on  the  surface  of  the  sphere ;  for,  the  three  points 
determine  a  plane  which  cuts  the  sphere  in  a  circle. 


PKOPOSITION  VIII.— THEOREM. 

35.  All  the  points  in  the  circumference  of  a  circle  of  the  sphere  are 
equally  distant  from  each  of  its  poles. 

Let  abed  be  any  circle  of  the  sphere  and 
PP'  the  diameter  of  the  sphere  perpendicu 
lar  to  its  plane ;  then,  by  the  definition  (27), 
Pand  P'  are  the  poles  of  the  circle  abed. 

Since  PP'  passes  through  the  centre  o 
of  the  circle,  the  distances  Pa,  Pb,  PC,  are 
oblique  lines  from  P  to  points  a,  b,  c,  equally 


BOOK    vin.  247 

distant  from  the  foot  of  the  perpendicular,  and  are  therefore  equal 
(VI.  10).  Hence,  all  the  points  of  the  circumference  abed  are  equally 
distant  from  the  pole  P.  For  the  same  reason,  they  are  equally  dis 
tant  from  the  pole  P'. 

36.  Corollary  I.  All  the  arcs  of  great  circles  drawn  from  a  pole 
of  a  circle  to  points  in  its  circumference,  as  the  arcs  Pa,  Pb,  PC,  are 
equal,  since  their  chords  are  equal  chords  in  equal  circles. 

By  the  distance  of  two  points  on  the  surface  of  a  sphere  is  usually 
understood  the  arc  of  a  great  circle  joining  the  two  points.  The 
arc  of  a  great  circle  drawn  from  any  point  of  a  given  circle  abc,  to 
one  of  its  poles,  as  the  arc  Pa,  is  called  the  polar  distance  of  the  given 
circle,  and  the  distance  from  the  nearest  pole  is  usually  understood. 

37.  Corollary  II.  The  polar  distance  of  a  great  circle  is  a  quad 
rant  of  a  great  circle;  thus  PA,  PB,  etc.,  P'A,  P'B,  etc.,  polar  dis 
tances  of  the  great  circle  ABCD,  are  quadrants;  for,  they  are  the 
measures  of  the  right  angles  AOP,  BOP,  AOP',  BOP',  etc.,  whose 
vertices  are  at  the  centre  of  the  great  circles  PAP',  PBP',  etc. 

In  connection  with  the  sphere,  by  a  quadrant  is  usually  to  be 
understood  a  quadrant  of  a  great  circle. 

38.  Corollary  III.  If  a  point  P  on  the  surface  of  the  sphere  is  at 
the  distance  of  a  quadrant  from  two  points,  B  and  C,  of  an  arc  of  a 
great  circle,  it  is  the  pole  of  that  arc.     For,  the  arcs  PB  and  PC 
being  quadrants,  the  angles  POB  and  POC  are  right  angles ;  there 
fore,  the  radius  OP  is  perpendicular  to  each  of  the  lines  OB,  OC, 
and   is   consequently  perpendicular   to   the   plane   of  the   arc  BC 
(VI.  13)  ;  hence,  Pis  the  pole  of  the  arc  BC. 

39.  Scholium.  By  means  of  poles,  arcs  of   circles  may  be  drawn 
upon  the  surface  of  a  sphere  with  the  same  ease  as  upon  a  plane  sur 
face.     Thus,  by  revolving  the  arc  Pa  about  the  pole  P,  its  extremity 
a  will  describe  the  small  circle  abd ;  and  by  revolving  the  quadrant 
PA  about  the  pole  P,  the  extremity  A  will  describe  the  great  circle 
ABD. 

If  two  points,  B  and  C,  are  given  on  the  surface,  and  it  is  required 
to  draw  the  arc  BC,  of  a  great  circle,  between  them,  it  will  be  neces 
sary  first  to  find  the  pole  P  of  this  circle ;  for  which  purpose,  take 
B  and  C  as  poles,  and  at  a  quadrant's  distance  describe  two  arcs  on 
the  surface  intersecting  in  P.  The  arc  B  C  can  then  be  described 
with  a  pair  of  compasses,  placing  one  foot  of  the  compasses  on  P  and 


248 


GEOMETRY. 


tracing  the  arc  with  the  other  foot.  The  opening  of  the  compasses 
(distance  between  their  feet)  must  in  this  case  be  equal  to  the  chord 
of  a  quadrant ;  and  to  obtain  this  it  is  necessary  to  know  the  radius 
of  the  sphere. 


PROPOSITION  IX.— PROBLEM. 


40.   To  find  the  radius  of  a  given  sphere. 

We  here  suppose  that  a  material  sphere  is  given,  and  that  only 
measurements  on  the  surface  are  possible. 


1st.  With  any  point  P  (Fig.  1)  of  the  given  surface  as  a  pole,  and 
with  any  arbitrary  opening  of  the  compasses,  describe  a  circum 
ference  abc  on  the  surface.  The  rectilinear  distance  Pa,  being  the 
arbitrary  opening  of  the  compasses,  is  a  known  line. 

Take  any  three  points,  a,  b,  c,  in  this  circumference,  and  with  the 
compasses  measure  the  rectilinear  distances  ab,  be,  ca. 

2d.  On  a  plane  surface  construct  a  triangle  abe  (Fig.  2),  with  the 
three  distances  ab,  bo,  ca,  and  find  the  centre  o  of  the  circle  circum 
scribed  about  the  triangle  (II.  87).  The  radius  ao  of  this  circle  is 
the  radius  of  the  circle  abc  of  Fig.  1. 

3d.  With  the  radius  ao  as  a  side,  and  the  known  distance  Pa  as 
the  hypotenuse,  construct  a  right  triangle  aoP  (Fig.  3).  Draw  aP' 
perpendicular  to  aP,  meeting  Po  produced  in  P'.  Then  it  is  evident 
that  PPf,  thus  determined,  is  equal  to  the  diameter  of  the  given 
sphere,  and  its  half  PO  is  the  required  radius. 

41.  Definition.  A  plane  is  tangent  to  a  sphere  when  it  has  but  one 
point  in  common  with  the  surface  of  the  sphere. 

42.  Definition.  Two  spheres  are  tangent  to  each  other  when  their 
surfaces  have  but  one  point  in  common. 


BOOK  viii.  249 


PROPOSITION  X.— THEOREM. 

43.  A  plane  perpendicular  to  a  radius  of  a  sphere  at  its  extremity  is 
tangent  to  the  sphere. 

Let  0  be  the  centre  of  a  sphere,  and 
let  the  plane  MN  be  perpendicular  to  a 
radius  OA  at  its  extremity  A ;  then,  the 
plane  MNis  tangent  to  the  sphere  at  the 
point  A. 

For,  taking  any  other  point,  as  H,  in 
the  plane,  and  joining  OH,  the  oblique 
line  OH  is  greater  than  the  perpendicu 
lar  OA  ;  therefore  the  point  H  is  without  the  sphere.     Hence  the 
plane  MN  has  but  the  point  A  in  common  with  the  sphere,  and  is 
consequently  tangent  to  the  sphere. 

44.  Corollary.  Conversely,  a  plane  tangent  to  a  sphere  is  perpen 
dicular  to  the  radius  drawn  to  the  point  of  contact.     For,  since  every 
point  of  the  plane  except  the  point  of  contact  is  without  the  sphere, 
the  radius  drawn  to  the  point  of  contact  is  the  shortest  line  from  the 
centre  of  the  sphere  to  the  plane,  therefore  it  is  perpendicular  to  the 
plane  (VI.  9). 

45.  Scholium.  Any  straight  line  A  T,  drawn  in  the  tangent  plane 
through  the  point  of  contact,  is  tangent  to  the  sphere. 

Any  two  straight  lines,  AT,  A  T',  tangent  to  the  sphere  at  the 
same  point  A,  determine  the  tangent  plane  at  that  point. 


PROPOSITION  XI.— PROBLEM. 

46.  Through  a  given  straight  line  without  a  given  sphere,  to  pass  a 
plane  tangent  to  the  sphere. 

Through  the  given  straight  line  and  the  centre  of  the  sphere,  a 
plane  can  be  passed  which  will  cut  the  sphere  in  a  great  circle.  Let 
the  plane  of  the  paper  represent  this  plane ;  let  MN  be  the  given 
line,  0  the  centre  of  the  sphere,  and  aPcP'  the  great  circle  in  which 
the  plane  passed  through  MN  and  the  centre  0  cuts  the  sphere. 

From  any  point  M  in  the  given  line  draw  a  tangent  MaT  to  the 
great  circle  aPc ;  draw  M 0  cutting  the  circumference  of  the  circle 


250 


G  E  O  M  E  T  K  Y. 


in  P  and  P';  let  fall  ao  perpen 
dicular  to  MO,  and  join  Oa. 

Conceive  the  sphere  to  be  gen 
erated  by  the  revolution  of  the 
semicircle  PaP'  about  its  diame 
ter,  and  let  the  tangent  Ma  re 
volve  with  it.  The  line  ao,  per 
pendicular  to  the  axis,  will  gener 
ate  a  small  circle  abc  whose  poles 
are  P  and  P' ;  the  tangent  MaT 
will  generate  a  conical  surface ; 

and  the  portion  of  this  surface  between  the  point  If  and  the  circum 
ference  abe  is  the  surface  of  the  cone  whose  vertex  is  M  and  whose 
base  is  the  circle  abc.  Every  element  of  this  cone  as  Mb  is  a  tangent 
to  the  sphere,  since  it  has  the  point  b,  and  that  point  only,  in  common 
with  the  sphere. 

Now,  every  plane  which  is  tangent  to  this  cone  is  also  tangent  to 
the  sphere ;  for  any  plane  touching  the  cone  in  an  element  Mb,  has 
the  point  b,  and  only  the  point  b,  in  common  with  the  sphere. 

Therefore  the  solution  of  the  present  problem  is  reduced  to  passing 
a  plane  through  the  given  line  MN,  tangent  to  the  cone  M-abc; 
which  is  done  by  Proposition  VI.  of  this  Book,  observing  the  Scho 
lium  (24). 

Since  there  are  two  tangent  planes  to  the  cone,  there  are  also  two 
tangent  planes  to  the  sphere,  passing  through  the  given  line  MN. 

47.  Scholium.  The  indefinite  conical  surface  generated  by  the 
revolution  of  the  tangent  MT  is  circumscribed  about  the  sphere ;  and 
the  sphere  is  inscribed  in  this  surface.  The  circle  abc  is  called  the 
circle  of  contact  of  the  cone  and  sphere. 


PKOPOSITION  XII.— THEOEEM. 

48.  The  intersection  of  two  spheres  is  a  circle  whose  plane  is  perpen 
dicular  to  the  straight  line  joining  the  centres  of  the  spheres,  and  whose 
centre  is  in  that  line. 

Through  the  centres  0  and  0'  of  the  two  spheres,  let  any  plane 
be  passed,  cutting  the  spheres  in  great  circles  which  intersect  each 
other  in  the  points  A  and  B ;  the  chord  AB  is  bisected  at  C  by  the 


BOOK     VIII. 


251 


line  00'  at  right  angles  (II.  34).  If  we 
now  revolve  the  plane  of  these  two  circles 
about  the  line  00',  the  circles  will  gener 
ate  the  two  spheres,  and  the  point  A  will 
describe  the  line  of  intersection  of  their 
surfaces.  Moreover,  since  the  line  AC 

will,  during  this  revolution,  remain  perpendicular  to  00',  it  will 
generate  a  circle  whose  plane  is  perpendicular  to  00'  (VI.  15),  and 
whose  centre  is  C. 

49.  Scholium.  Two  spheres  being  given  in  any  position  whatever, 
if  any  plane  is  passed  through  their  centres  cutting  them  in  two 
great  circles,  the  spheres  will  intersect  if  these  circles  intersect,  will 
be  tangent  to  each  other  if  these  circles  are  tangent  to  each  other, 
etc.  For  each  of  these  positions,  therefore,  we  shall  have  the  same 
relations  between  the  distance  of  the  centres  and  the  radii  of  the 
spheres,  as  have  been  established  for  the  corresponding  positions  of 
two  circles  in  Book  II. 


PEOPOSITION  XIII.— THEOKEM. 

50.  Through  any  four  points  not  in  the  same  plane,  a  spherical  sur 
face  can  be  made  to  pass,  and  but  one. 

Let  A,  B,  C,  D,  be  four  given  points  not 
in  the  same  plane.  These  four  points  may 
be  taken  as  the  vertices  of  a  tetraedron 
ABCD. 

Let  E  be  the  centre  of  the  circle  circum 
scribed  about  the  face  ABC,  and  draw  EM 
perpendicular  to  this  face;  every  point  in 
EM  is  equally  distant  from  the  points  A,  B 
and  C  (VI.  10). 

Let  F  be  the  centre  of  the  circle  circum 
scribed  about  the  face  BCD,  and  draw  FN  perpendicular  to  this 
face ;  every  point  in  FN  is  equally  distant  from  the  points  B,  C 
andD. 

The  two  perpendiculars,  EM  and  FN,  intersect  each  other.  For, 
let  H  be  the  middle  point  of  BC,  and  draw  EH,  FH.  The  lines 
EH  and  FH  are  each  perpendicular  to  BC  (II.  16);  therefore, the 


252  GEOMETRY. 

plane  passed  through  EH  and.  FH  is  perpendicular  to  BC  (VI.  13) 
and  consequently  also  to  each  of  the  faces  ABC,  BCD  (VI.  47). 
Hence,  the  perpendiculars  EM  and  FN  lie  in  the  same  plane  EHF 
(VI.  50),  and  must  meet  unless  they  are  parallel ;  but  they  cannot  be 
parallel  unless  the  planes  BCD  and  ABC  are  one  and  the  same 
plane,  which  is  contrary  to  the  hypothesis  that  the  four  given  points 
are  not  in  the  same  plane. 

The  intersection  0  of  the  perpendiculars  EM  and  FN,  being 
equally  distant  from  A,  B  and  C,  and  also  equally  distant  from  B, 
C  and  D,  is  equally  distant  from  the  four  points  A,  B,  C  and  D ; 
therefore,  a  spherical  surface  whose  centre  is  0  and  whose  radius  is 
the  distance  of  0  from  any  one  of  these  points,  will  pass  through 
them  all. 

Moreover,  since  the  centre  of  any  spherical  surface  passing  through 
the  four  points  A,  B,  C  and  D  is  necessarily  in  each  of  the  perpen 
diculars  EM,  FN,  the  intersection  0  is  the  centre  of  the  only  spheri 
cal  surface  that  can  be  made  to  pass  through  the  four  given 
points. 

51.  Corollary  I.  The  four  perpendiculars  to  the  planes  of  the  faces 
of  a  tetraedron,  erected  at  the  centres  of  the  faces,  meet  in  the  same 
point. 

52.  Corollary  II.  The  six  planes,  perpendicular  to  the  six  edges 
of  a  tetraedron  at  their  middle  points,  intersect  in  the  same  point. 


PKOPOSITION  XIV.— THEOREM. 

53.  A  sphere  may  be  inscribed  in  any  given  tetraedron. 

Let  ABCD  be  the  given  tetraedron. 

Let  the  planes  OAB,  OBC,  OAC,  bisect  the 
diedral  angles  at  the  edges  AB,  BC,  AC,  re 
spectively.  Every  point  in  the  plane  OAB  is 
equally  distant  from  the  faces  ABC  and  ABD 
(VI.  55);  every  point  in  the  plane  OBC  is 
equally  distant  from  the  faces  ABC  and  DBC', 
and  every  point  in  the  plane  OAC  is  equally 
distant  from  the  faces  ABC  and  ADC',  there 
fore,  the  common  intersection,  0,  of  these  three  planes  is  equally 
distant  from  the  four  faces  of  the  tetraedron  ;  and  a  sphere  described 


BOOK     VIII. 


253 


with  0  as  a  centre,  and  with  a  radius  equal  to  the  distance  of  0  from 
any  face,  will  be  tangent  to  each  face,  and  will  be  inscribed  in  the 
tetraedron. 

54.  Corollary.  The  six  planes,  bisecting  the  six  diedral  angles  of  a 
tetraedron,  intersect  in  the  same  point. 


SPHEKICAL   ANGLES. 

55.  Definition.  The  angle  of  two  curves  passing  through  the  same 
point  is  the  angle  formed  by  the  two  tangents  to  the  curves  at  that 
point. 

This  definition  is  applicable  to  any  two  intersecting  curves  in 
space,  whether  drawn  in  the  same  plane  or  upon  a  surface  of  any 
kind. 

27 

Thus,  in  a  plane,  two  circumferences  inter 
secting  in  a  point  A,  make  an  angle  equal  to 
the  angle  TAT'  formed  by  their  tangents  at 
A.  In  this  case,  the  angle  is  also  equal  to 
the  angle  OAO'  formed  by  the  radii  of  the 
two  circles  drawn  to  the  common  point. 


In  like  manner,  on  a  sphere,  the  angle 
formed  by  any  two  intersecting  curves, 
AB,  AB',  is  the  angle  TAT',  formed  by 
the  lines  A T,  AT',  tangents  to  the  two 
curves,  respectively,  at  their  common 
point  A. 


PEOPOSITION  XV.— THEOKEM. 

56.   The  angle  of  two  intersecting  curves  on  the  surface  of  a  sphere  is 
equal  to  the  diedral  angle  between  the  planes  passed  through  the  centre 
of  the  sphere  and  the  tangents  to  the  two  curves  at  their  point  of  in 
tersection. 
22 


254 


GEOMETRY. 


Let  the  curves,  AB  and  AB',  on  the 
surface  of  a  sphere  whose  centre  is  0,  in 
tersect  at  A,  and  let  AT  and  AT'  be  the 
tangents  to  the  two  curves,  respectively. 
Since  J.Tand  AT'  do  not  cut  the  curves 
at  A,  they  do  not  cut  the  surface  of  the 
sphere,  and  are  therefore  tangents  to  the 
sphere.  Hence  they  are  both  perpendicular  to  the  radius  OA  drawn 
to  the  common  point  of  contact,  and  consequently  the  angle  T'A  T, 
which  is  the  angle  of  the  two  curves  (55),  measures  the  diedral  angle 
of  the  planes  OA  T,  OA  T',  passed  through  the  radius  OA  and  each 
of  the  tangents. 


PROPOSITION  XVI.— THEOREM. 

57.  The  angle  of  two  arcs  of  great  circles  is  equal  to  the  angle  of 
their  planes,  and  is  measured  by  the  arc  of  a  great  circle  described  from 
its  vertex  as  a  pole  and  included  between  its  sides  (produced  if  ne 
cessary'). 

Let  AB  and  AB'  be  two  arcs  of  great 
circles,  A  T  and  A  T'  the  tangents  to  these 
arcs  at  A,  0  the  centre  of  the  sphere. 
The  planes  passing  through  the  centre  0 
and  the  tangents  AT,  AT',  are  in  this 
case  the,  planes  of  the  curves  AB,  AB', 
themselves;  consequently  the  angle  BAB', 
or  TAT',  is  equal  to  the  angle  of  these 

planes  (56),  the  edge  of  this  angle  being  the  common  diameter 
AOD. 

Now  let  CC'  be  the  arc  of  a  great  circle  described  from  A  as  a 
pole  and  intersecting  the  arcs  AB,  AB'  (produced  if  necessary),  in 
C  and  C'.  The  radii  OC  and  OC'  are  perpendicular  to  AO,  since 
the  arcs  AC,  AC',  are  quadrants  (37) ;  therefore,  the  angle  COC'  is 
also  equal  to  the  diedral  angle  AO,  or  to  the  angle  BAB',  and  it  is 
measured  by  the  arc  CC'. 

58.  Corollary.  Any  great  circle  arc  AC',  drawn  through  the  pole 
of  a  given  great  circle  CC',  is  perpendicular  to  the  circumference 
CC'.     For,  the  pole  A  being  in  the  diameter  AOD  perpendicular  to 


BOOK   vin.  255 

the  plane  of  CC',  the  plane  of  AC'  is  perpendicular  to  the  plane  of 
CC'  (VI.  47),  and  hence  the  angle  C'  is  a  right  angle. 

Conversely,  any  great  circle  arc  C'A  perpendicular  to  the  arc  CC' 
must  pass  through  the  pole  A  of  CC'. 

59.  Scholium.  If  it  is  required  to  draw  a  great  circle  B '  C'  perpen 
dicular  to  a  given  great  circle  CC'E,  through  a  given  point  B',  we 
have  only  to  find  the  pole  N  of  the  required  arc  by  describing,  from 
Br  as  a  pole  and  at  a  quadrant's  distance,  an  arc  cutting  CC'E  in 
N-,  then,  from  N  as  a  pole,  the  perpendicular  B' C'  can  be  de 
scribed. 


SPHERICAL    POLYGONS    AND    PYRAMIDS. 

60.  Definition.  A  spherical  polygon  is  a  portion 
of  the  surface  of  a  sphere  bounded  by  three  or 
more  arcs  of  great  circles,  as  ABCD. 

Since  the  planes  of  all  great  circles  pass 
through  the  centre  of  the  sphere,  the  planes  of 
the  sides  of  a  spherical  polygon  form,  at  the  cen 
tre  0,  a  polyedral  angle  of  which  the  edges  are  the  radii  drawn  to 
the  vertices  of  the  polygon,  the  face  angles  are  angles  at  the  centre 
measured  by  the  sides  of  the  polygon,  and  the  diedral  angles  are 
equal  to  the  angles  of  the  polygon  (57). 

Since  in  a  polyedral  angle  each  face  angle  is  assumed  to  be  less 
than  two  right  angles,  each  side  of  a  spherical  polygon  will  be  as 
sumed  to  be  less  than  a  semi-circumference. 

A  spherical  polygon  is  convex  when  its  corresponding  polyedral 
angle  at  the  centre  is  convex  (VI.  67). 

A  diagonal  of  a  spherical  polygon  is  an  arc  of  a  great  circle  join 
ing  any  two  vertices  not  consecutive. 

61.  Definition.  A  spherical  triangle  is  a  spherical  polygon  of  three 
sides.     It  is  called  right  angled,  isosceles,  or  equilateral,  in  the  same 
cases  as  a  plane  triangle. 

62.  Definition.  A  spherical  pyramid  is  a  solid  bounded  by  a  spheri 
cal  polygon  and  the  planes  of  the  sides  of  the  polygon ;  as  0-AB  CD. 
The  centre  of  the  sphere  is  the  vertex  of  the  pyramid ;  the  spherical 
polygon  is  its  base. 


256 


GEOMETRY. 


63.  Symmetrical  spherical  triangles  and  polygons.     Let  ABC  be  a 
spherical  triangle,  and  0  the  centre  of  the 

sphere.  Drawing  the  radii  OA,  OB,  OC,  we 
form  the  triedral  angle  0-ABC,  at  the  centre. 
The  sides  AB,  BC,  AC,  of  the  triangle  are 
respectively  the  measures  of  the  face  angles 
AOB,BOC,AOC,  of  the  triedral  angle ;  and 
the  angles  A,  B,  C,  of  the  triangle  are  respec 
tively  equal  to  the  diedral  angles  at  the  edges 
OA,  OB,  OC,  of  the  triedral  angle  (57). 

If  the  radii  A 0,  BO,  CO,  are  produced  to  meet  the  surface  of  the 
sphere  in  the  points  A',  Br,  C',  and  if  these  points  are  joined  by  arcs 
of  great  circles  A'B',  B'C',  AC',  a  triedral  angle  0-A'B'C'  is 
formed  symmetrical  with  0-ABC  (VI.  68),  and  its  corresponding 
spherical  triangle  A'B1 ' Cr  is  symmetrical  with  ABC. 

The  spherical  pyramid  0-A'B'C'  is  also  symmetrical  with  the 
spherical  pyramid  0-ABC. 

In  the  same  manner,  we  may  form  two  symmetrical  polygons  of 
any  number  of  sides,  and  corresponding  symmetrical  pyramids. 

64.  Two  symmetrical  spherical  triangles,  or  polygons,  are  still 
called  symmetrical  in  whatever  position  they  may  be  placed  on  the 
surface  of  the  sphere.     If  we  place  the  symmetri- 

cal  triangles  of  the  preceding  figure  with  the  ver 
tices  A'  and  B'  in  coincidence  with  their  homolo 
gous  vertices  A  and  B,  their  third  vertices  C  and 
C'  will  lie  on  opposite  sides  of  the  arc  AB.  In 
this  position,  it  is  apparent  that  the  order  of  ar 
rangement  of  the  parts  in  one  triangle  is  the 
reverse  of  that  in  the  other,  and  that,  in  general, 
two  symmetrical  spherical  triangles  cannot  be  made  to  coincide  by 
superposition. 

65.  There  is,  however,  one  exception  to  the  last  remark,  namely, 
the   case   of  symmetrical  isosceles   tri 
angles.     For,  if  ABC  is  an  isosceles 

spherical  triangle  and  AB  =  AC, 
then,  in  its  symmetrical  triangle  we 
have  AB'  =  AC',  and  consequently 
AB  =A'C',AC  =  AB',  and  since 


BOOK    VIII. 


257 


the  angles  A  and  A'  are  equal,  if  AB  be  placed  on  A'  C',  A  C  will 
fall  on  its  equal  A'B'  and  the  two  triangles  will  coincide  throughout. 

66.  In  consequence  of  the  relation  established  between  polyedral 
angles  and  spherical  polygons,  it  follows  that  from  any  property  of 
polyedral  angles  we  may  infer  an  analogous  property  of  spherical 
polygons. 

Keciprocally,  from  any  property  of  spherical  polygons  we  may 
infer  an  analogous  property  of  polyedral  angles. 

The  latter  is  in  almost  all  cases  the  more  simple  mode  of  proce 
dure,  inasmuch  as  the  comparison  of  figures  drawn  on  the  surface  of 
a  sphere  is  nearly  if  not  quite  as  simple  as  the  comparison  of  plane 
figures. 

67.  Definition.   If  from  the  vertices  of  a  spherical   triangle  as 
poles,  arcs  of  great  circles  are  described,  these  arcs  form  by  their 
intersection  a  second  triangle  which  is  called  the  polar  triangle  of  the 
first. 

Thus,  if  A,  B  and  C  are  the  poles  of  the  arcs 
of  great  circles,  B'C',  AC',  and  A'B',  respec 
tively,  AB'C'  is  the  polar  triangle  of  ABC. 

Since  all  great  circles,  when  completed,  intersect 
each  other  in  two  points,  the  arcs  B'C',  AC', 
A'B',  if  produced,  will  form  three  other  triangles; 
but  the  triangle  which  is  taken  as  the  polar  tri 
angle  is  that  whose  vertex  A,  homologous  to  A,  lies  on  the  same 
side  of  the  arc  BC  as  the  vertex  A ;  and  so  of  the  other  vertices. 


PKOPOSITION  XVII.— THEOKEM. 

68.  If  AB'C'  is  the  polar  triangle  of  ABC,  then,  reciprocally, 
ABC  is  thepolar  triangle  of  AB'C'. 

For,  since  A  is  the  pole  of  the  arc  B' C',  the 
point  B '  is  at  a  quadrant's  distance  from  A  ;  and 
since  C  is  the  pole  of  the  arc  A'B',  the  point  B'  is 
at  a  quadrant's  distance  from  C;  therefore,  B'  is 
the  pole  of  the  arc  A  C  (38).  In  the  same  man 
ner,  it  is  shown  that  A  is  the  pole  of  the  arc  BC, 
and  C'  the  pole  of  the  arc  AB.  Moreover,  A  and 
A  are  on  the  same  side  of  B'  C',  B  and  B'  on  the  same  side  of  AC', 

22*  R 


258 


GEOMETRY. 


C  and  C'  on  the  same  side  of  A'B1 ;  therefore,  JJ?C  is  the  polar 
triangle  of  A'B'C'. 


PKOPOSITION  XVIII.— THEOREM. 

69.  In  two  polar  triangles,  each  angle  of  one  is  measured  by  the  sup 
plement  of  the  side  lying  opposite  to  it  in  the  other. 

Let  ABC  and  A'B' C'  be  two  polar  triangles. 

Let  the  sides  AB  and  A  C,  produced  if  necessary, 
meet  the  side  B' C'  in  the  points  b  and  c.  The 
vertex  A  being  the  pole  of  the  arc  be,  the  angle 
A  is  measured  by  the  arc  be  (57). 

Now,  Br  being  the  pole  of  the  arc  Ac  and  C' 
the  pole  of  the  arc  Ab,  the  arcs  B'c  and  C'b  are 
quadrants ;  hence  we  have 

B'C'  +  be  =  B'e  -f  C'b  =  a  semi-circumference. 

Therefore  be,  which  measures  the  angle  A,  is  the  supplement  of  the 
sideJB'C"(IL55). 

In  the  same  manner,  it  can  be  shown  that  each  angle  of  either 
triangle  is  measured  by  the  supplement  of  the  side  lying  opposite  to 
it  in  the  other  triangle. 

70.  Scholium  I.  Let  the  angles  of  the  triangle 
ABC  be  denoted  by  A,  B  and  C,  and  let  the  sides 
opposite  to  them,  namely,  BC,  AC  and  AB,  be 
denoted  by  a,  b  and  c,  respectively.     Let  the  cor-  f 
responding  angles  and  sides  of  the  polar  triangle 
be  denoted  by  A,  B',  C',  a',  b'  and  c'.     Also  let 
both  angles  and  sides   be  expressed   in  degrees 

(II.  54).    Then,  the  preceding  theorem  gives  the  following  relations : 

A  +  a'  =  B  +  b'  =  C  +  c'  =  180°, 
A'  +  a  =B'  +  b  =  C'  +  c  =  180°, 

also  A  —  a  =  A'  —  a',  etc. 

71.  Scholium  II.  Two  triedral  angles  at  the  centre  of  the  sphere, 
corresponding  to  two  polar  triangles  on  the  surface,  are  called  sup 
plementary    triedral   angles;    for,    it    follows    from    the    preceding 


BOOK   viii.  259 

theorem,  and  from  the  relation  between  any  spherical  polygon  and 
its  corresponding  polyedral  angle  (60),  that  the  diedral  angles  of 
either  of  these  triedral  angles  are  respectively  the  supplements  of 
the  opposite  face  angles  of  the  other. 

PROPOSITION  XIX.— THEOEEM. 

72.  Two  triangles  on  the  same  sphere  are  either  equal  or  symmetrical, 
when  two  sides  and  the  included  angle  of  one  are  respectively  equal  to 
two  sides  and  the  included  angle  of  the  other. 

In  the  triangles  ABC  and  DEF,  let  the  angle 
A  be  equal  to  the  angle  D,  the  side  AB  equal 
to  the  side  DE,  and  the  side  A  C  equal  to  side 
DF. 

1st.  When  the  parts  of  the  two  triangles  are 
in  the  same  order,  ABC  can  be  applied  to 
DEF,  as  in  the  corresponding  case  of  plane 
triangles  (I.  76),  and  the  two  triangles  will 
coincide ;  therefore,  they  are  equal. 

2d.  When  the  parts  of  the  two  tri- 
angles  are  in  inverse  order,  let  DE  'F  be 
the  symmetrical  triangle  of  DEF,  and 
therefore  having  its  angles  and  sides  equal, 
respectively,  to  those  of  DEF.  Then,  in 
the  triangle  ABC  and  DE'F,  we  shall 
have  the  angle  BA  C  equal  to  the  angle 
E'DF,  the  side  AB  to  the  side  DE',  and 
the  side  A  C  to  the  side  DF,  and  these  parts  arranged  in  the  same 
order  in  the  two  triangles;  therefore,  the  triangle  ABC  is  equal  to 
the  triangle  DE'F,  and  consequently  symmetrical  with  DEF. 

73.  Scholium.  In  this  proposition,  and  in  those  which  follow,  the 
two  triangles  may  be  supposed  on  the  same  sphere,  or  on  two  equal 
spheres. 

PROPOSITION  XX.— THEOEEM. 

74.  Two  triangles  on  the  same  sphere  are  either  equal  or  symmetrical, 
when  a  side  and  the  two  adjacent  angles  of  one  are  equal  respectively  to 
a  side  and  the  two  adjacent  angles  of  the  other. 


260  GEOMETRY. 

For,  one  of  the  triangles  may  be  applied  to  the  other,  or  to  its 
symmetrical  triangle,  as  in  the  corresponding  case  of  plane  tri 
angles  (I.  78). 

PKOPOSITION  XXI.— THEOREM. 

75.  Two  triangles  on  the  same  sphere  are  either  equal  or  symmetrical, 
when  the  three  sides  of  one  are  respectively  equal  to  the  three  sides  of  the 
other. 

For,  their  corresponding  triedral  angles  at  the  centre  of  the  sphere 
are  either  equal  or  symmetrical  (VI.  71). 


PROPOSITION  XXII.— THEOREM. 

76.  If  two  triangles  on  the  same  sphere  are  mutually  equiangular, 
they   are   also    mutually   equilateral;   and   are   either   equal  or  sym 
metrical. 

Let  the  spherical  triangles 
M  and  N  be  mutually  equian 
gular. 

Let  M '  be  the  polar  triangle 
of  M,  and  -ZV'  the  polar  triangle  of  N.  Since  M  and  N  are  mutually 
equiangular,  their  polar  triangle's  M'  and  N'  are  mutually  equi 
lateral  (69)  ;  therefore,  by  the  preceding  proposition,  the  triangles  M' 
and  N'  are  mutually  equiangular.  But  M'  and  N'  being  mutually 
equiangular,  their  polar  triangles  M  and  N  are  mutually  equilateral 
(69).  Consequently,  M  and  N  are  either  equal  or  symmetrical  (75). 

77.  Scholium.  It   may  seem   to   the   student   that   the   preceding 
property  destroys  the  analogy  which   subsists  '  between  plane  and 
spherical  triangles,  since  two  mutually  equiangular  plane  triangles 
are  not  necessarily  mutually  equilateral.     But  in  the  case  of  spheri 
cal  triangles,  the  equality  of  the  sides  follows  from  that  of  the  angles 
only  upon  the  condition  that  the  triangles  are  constructed  upon  the 
same  sphere  or  on  equal  spheres ;  if  they  are  constructed  on  spheres 
of  different  radii,  the  homologous  sides  of  two  mutually  equiangular 
triangles  will  no  longer  be  equal,  but  will  be  proportional  to  the 
radii  of  the  sphere ;  the  two  triangles  will  then  be  similar,  as  in  the 
case  of  plane  triangles. 


BOOK    VIII.  261 


PKOPOSITION  XXIIL— THEOKEM. 

78.  In  an  isosceles  spherical  triangle,  the  angles  opposite  the  equal 
sides  are  equal. 

In  the  spherical  triangle  ABC,  let  AB  =  AC',  A 

then,  B  =  C. 

For,  draw  the  arc  AD  of  a  great  circle,  from  the 
vertex  A  to  the  middle  of  the  base  BC.  The  tri 
angles  ABD  and  A  CD  are  mutually  equilateral, 
and  in  this  case  are  symmetrical  (77) ;  therefore 

B  =  a 

79.  Corollary.  Since  the  triangles  ABD  and  ACD  are  mutually 
equiangular,  we  have  the  angle  BAD  equal  to  the  angle  CAD,  and 
the  angle  ADB  equal  to  the  adjacent  angle  ADC',  therefore,  the  arc 
drawn  from  the  vertex  of  an  isosceles  spherical  triangle  to  the  middle  of 
the  base  is  perpendicular  to  the  base  and  also  bisects  the  vertical  angle. 

80.  Scholium.    This  proposition   and   its   corollary  may  also   be 
proved  by  applying  the   isosceles  triangle  to  its  symmetrical  tri 
angle  (65). 


PKOPOSITION  XXIV.— THEOKEM. 

81.  If  two  angles  of  a  spherical  triangle  are  equal,  the  triangle  is 
isosceles. 

In  the  triangle  ABC  let  B  —  C',  then, 
AB  =  AC. 

For,  letA'B'C'  be  the  polar  triangle  of  ABC. 
Then,  the  sides  A'B'  and  A'C'  are  equal 
(69),  and  therefore  the  angles  B '  and  C'  are 
equal  (78).  But  since  the  angles  B'  and  C" 
are  equal  in  the  triangle  A'B'C',  the  sides  AB 
and  AC  are  equal  in  its  polar  triangle  ABC. 


262  GEOMETRY. 


PROPOSITION  XXV.— THEOREM. 

82.  Any  side  of  a  spherical  triangle  is  less  than  the  sum  of  the 
other  two. 

Let  ABC  be  a  spherical  triangle;  then, 
any  side,  as  A  C,  is  less  than  the  sum  of  the 
other  two,  AS  and  AC. 

For,  in  the  corresponding  triedral  angle 
formed  at  the  centre  0  of  the  sphere,  we 
have  the  angle  AOC  less  than  the  sum  of 
the  angles  AOB  and  BOO  (VI.  69);  and 

since  the  sides  of  the  triangle  measure  these  angles,  respectively,  we 
have  AC<AB  -{-  BC. 

83.  Corollary.   Any  side,  AB,  of  a  spherical 
polygon  ABCDEis  less  than  the  sum  of  all  the 
other  sides. 

PROPOSITION  XXVI.— THEOREM. 

84.  In  a  spherical  triangle,  the  greater  side  is  opposite  the  greater 
angle ;  and  conversely. 

1st.  In  the  triangle  ABC  suppose  A  B  C  >  A  CB ;  A 
then,  AC>  AB.  For,  draw  the  arc  BD  making 
the  angle  DBC=DCB;  then,  the  triangle  BDC 
is  isosceles  (81),  and  DC=DB.  Adding  DA  to 
each  of  these  equals  we  have  AC  =  DB  -f-  DA. 
But  DB  +  DA>  AB  (82)  ;  therefore,  AC>  Aff. 

2d.  Conversely,  in  the  triangle  AB  C  suppose  A  C  >  A  B ;  then 
ABC>  ACB.  For,  if  ABC  were  equal  to  ACB,  AC  would  be 
equal  to  AB  (81),  which  is  contrary  to  the  hypothesis  ;  and  if  ABC 
were  less  than  A  CB,  A  C  would  be  less  than  AB,  which  is  also  con 
trary  to  the  hypothesis;  therefore,  ABC  must  be  greater  than  ACB, 

PROPOSITION  XXVII.— THEOREM. 

85.  If  from  the  extremities  of  one  side  of  a  spherical  triangle  two  arcs 
of  great  circles  are  drawn  to  a  point  within  the  triangle,  the  sum  of 
these  arcs  is  less  than  the  sum  of  the  other  two  sides  of  the  triangle. 


BOOK    viii.  263 

In  the  spherical  triangle  ABC,  let  the  arcs 
BD  and  CD  be  drawn  to  any 'point  D  within  the 
triangle ;  then,  DE  -f  DC  <  AB  +  A C. 

For,  produce  BD  to  meet  AC  in  E;  then  we 
have  DC  <  DE  +  EC  (82)  ;  and  adding  BD  to 
both  members  of  this  inequality,  we  have  DB  -f  DC  <  BE  -(-  EC. 
In   the  same  manner,   we   prove   that  BE  -f-  EC  <  AB  -J-  AC; 
therefore,  DB  -{-  DC<AB  +  AC. 

PKOPOSITION  XXVIII.— THEOKEM. 

86.  The  sum  of  the  sides  of  a  convex  spherical  polygon  is  less  than 
the  circumference  of  a  great  circle. 

For,  the  sum  of  the  face  angles  of  the  corresponding  polyedral 
angle  at  the  centre  of  the  sphere  is  less  than  four  right  angles 
(VI.  70). 

PKOPOSITION  XXIX.— THEOEEM. 

87.  The  sum  of  the  angles  of  a  spherical  triangle  is  greater  than 
two,  and  less  than  six,  right  angles. 

For,  denoting  the  angles  of  a  spherical  triangle 
by  A,  B,  C,  and  the  sides  respectively  opposite  to 
them  in  its  polar  triangle  by  a',b',cr,  we  have  (70), 

A  =  180°  —  a',  B  =  180°  —  b',C=  180°  —  e', 
the  sum  of  which  is 

A  +  B  +  C  =  540°  —  (a'  +  V  +  e'). 
But  a'  +  br  -f  c'  <  360°  (86) ;  therefore,  A  -f  B  -f  C  >  180°; 
that  is,  the  sum  of  the  three  angles  is  greater  than  two  right  angles. 
Also,  since  each  angle  is  less  than  two  right  angles,  their  sum  is  less 
than  six  right  angles. 

88.  Corollary.  A  spherical  triangle  may  have  two  or  even  three 
right  angles ;  also  two  or  even  three  obtuse  angles. 

89.  Definitions.   If  a  spherical  triangle  ABC  has 
two  right  angles,  I?  and  C,  it  is  called  a  bi-rectangular 
triangle ;  and  since  the  sides  AB  and  A  C  must  each 
pass  through  the  pole  of  BC  (58),  the  vertex  A  is 
that  pole,  and  therefore  AB  and  AC  are  quadrants. 


264 


GEOMETRY. 


If  a  triangle  has  three  right  angles  it  is 
called  a  tri-rectangular  triangle;  each  of 
its  sides  is  a  quadrant,  and  each  vertex  is 
the  pole  of  the  opposite  side.  Three  planes 
passed  through  the  centre  of  the  sphere, 
each  perpendicular  to  the  other  two  (VI.  48), 
divide  the  surface  of  the  sphere  into  eight 
tri-rectangular  triangles,  ABC,  ABC,  etc. 


EATIO    OF    THE    SUEFACES    AND  VOLUMES    OF    SPHEEICAL 

FIGUEES. 

90.  Definitions.  A  lune  is  a  portion  of  the  surface 
of  a  sphere  included  between  two  semi-circumferences 
of  great  circles ;  as  AMBNA. 

A  spherical  ungula,  or  wedge,  is  a  solid  bounded  by 
a  lune  and  the  two  semicircles  which  intercept  the 
lune  on  the  surface  of  the  sphere ;  as  the  solid 
ABMANB.  The  common  diameter  AB,  of  the  semi 
circles,  is  called  the  edge  of  the  ungula;  the  lune  is  called  its 
base. 

91.  Definition.  The  excess  of  the  sum  of  the  angles  of  a  spherical 
triangle  over  two  right  angles  is  called  the  spherical  excess. 

If  the  angles  of  a  spherical  triangle  ABC  are  denoted  by  A,  B 
and  C,  and  its  spherical  excess  by  E,  and  if  a  right  angle  is  the  unit 
employed  in  expressing  the  angles,  we  shall  have 

E=A+B+C—  2. 


PEOPOSITION  XXX.— THEOEEM. 

92.   Two  symmetrical  spherical  triangles  are  equivalent. 

Let  ABC  and  A'B'C'be  two  symmetrical  triangles  with  their 
homologous  vertices  diametrically  opposite  to  each  other  on  the 
sphere.  Let  P  be  the  pole  of  the  small  circle  which  passes  through 
the  three  points  A,  B  and  C.  The  great  circle  arcs  PA,  PB,  PC, 
are  equal  (36). 


BOOK     VIII. 


265 


Draw  the  diameter  POP'  and  the  great 
circle  arcs  P'A',  P'B',  P'C";  these  arcs 
being  equal,  respectively,  to  PA,  PB,  PC, 
are  also  equal  to  each  other. 

The  triangles  PAB,  P'A'B',  are  mu 
tually  equilateral,  and  also  isosceles; 
therefore,  they  are  superposable  (65)  and 
are  equal  in  area.  For  the  same  reason 
the  triangle  PA  C  is  equivalent  to  the  tri 
angle  P'A'C',  and  PBC  is  equivalent  to  P'B'C'.  Therefore  the 
triangle  ABC,  which  is  the  sum  of  the  triangles  PAB,  PAC  and 
PBC,  is  equivalent  to  its  symmetrical  triangle  A'B' C'  which  is  the 
sum  of  the  triangles  P'A'B',  P'A'C'  and  P'B'C'. 

If  the  pole  P  should  fall  without  the  triangle  ABC,  the  triangle 
would  be  equivalent  to  the  sum  of  two  of  the  isosceles  triangles 
diminished  by  the  third  ;  but  as  the  same  thing  would  occur  for  the 
symmetrical  triangle,  the  conclusion  would  be  the  same. 

93.  Corollary  I.  If  the  arcs  of  two  great 
circles,  AC  A,  BCB',  intersect  on  the  sur 
face  of  a  hemisphere,  the  sum  of  the  oppo 
site  triangles  ACB,  A CB',  is  equivalent  to 
a   lune  whose   angle   is   the  angle  ACB, 
formed  by  the  great  circles. 

For,  completing  the  great  circle  BCB'C', 
the  triangles  A'CB',  AC'B,  are  symmetri 
cal,  and  therefore  equivalent.     Hence,  the  sum  of  ACB  and  A'CB' 
is  equivalent  to  the  sum  of  A  CB  and  A  C'B,  that  is,  to  the  lune 
ACBC'A,  whose  angle  is  the  angle  ACB. 

94.  Corollary  II.  The  reasoning  employed  in  the  demonstration 
of  the  theorem  may  be  applied  also  to  the  pyramids  whose  bases  are 
two  symmetrical  triangles.     Hence,  two  symmetrical  spherical  triangu 
lar  pyramids  are  equivalent. 

Also  by  the  reasoning  in  Corollary  I.  we  infer  that  the  sum  of  the 
volumes  of  two  spherical  triangular  pyramids  the  sum  of  whose  bases  is 
equivalent  to  a  lune,  is  equal  to  the  volume  of  the  ungula  whose  base  is 
that  lune. 
23 


266  GEOMETRY. 


PROPOSITION  XXXI.—  THEOREM. 

95.  A  lune  is  to  the  surface  of  the  sphere  as  the  angle  of  the  lune  is  to 
four  right  angles. 

Let  ANBMA  be  a  lune,  and  let  MNP  be 
the  great  circle  whose  poles  are  the  ex 
tremities  of  the  diameter  AB. 

Let  the  circumference  of  the  circle  MNP 
be  divided  into  any  number  of  equal  parts 
Ma,  ab,  etc.  ;  and  let  planes  be  passed 
through  the  diameter  AB  and  each  of  the 
points  of  division.  The  whole  surface  of 

the  sphere  will  evidently  be  divided  into  equal  lunes  of  which  the 
given  lune  will  contain  the  same  number  as  there  are  parts  in  the 
arc  MN.  Hence,  whether  the  number  of  the  parts  in  MN  and  the 
number  of  the  parts  in  the  whole  circumference  MNP,  are  commen 
surable  or  incommensurable,  the  ratio  of  the  lune  ANBMA  to  the 
surface  of  the  sphere  is  the  same  as  the  ratio  of  the  arc  MNto  the 
circumference  MNP  ;  or,  since  MN  is  the  measure  of  the  angle  of 
the  lune,  and  the  circumference  MNP  is  the  measure  of  four  right 
angles,  the  lune  is  to  the  surface  of  the  sphere  as  the  angle  of  the 
lune  is  to  four  right  angles. 

96.  Corollary  I.  Two  lunes,  on  the  same  or  on  equal  spheres,  are 
to  each  other  as  their  angles. 

97.  Corollary  II.  If  we  denote  the  surface  of  the  tri-rectangular 
triangle  by  T,  the  surface  of  the  whole  sphere  'will  be  ST  (89); 
therefore,  denoting  the  surface  of  the  lune  by  L  and  its  angle  by  A, 
the  unit  of  the  angle  being  a  right  angle,  we  have 

—  =  —  ,  whence  L  =  T  X  2A. 

If,  further,  we  take  the  tri-rectangular  triangle  as  the  unit  of  sur 
face  in  comparing  surfaces  on  the  same  sphere,  we  shall  have 


that  is,  a  right  angle  being  the  unit  of  angles,  and  the  tri-rectangular 


BOOK  viii.  267 

triangle  the  unit  of  spherical  surfaces,  the  area  of  a  lune  is  expressed  by 
twice  its  angle. 

98.  Corollary  III.  The  tri-rectangular  spherical  pyramid  (that 
whose  base  is  the  tri-rectangular  triangle)  being  taken  as  the  unit  of 
volume,  the  same  reasoning  may  be  employed  to  prove  that  the 
volume  of  an  ungula  will  be  expressed  by  twice  its  angle. 


PKOPOSITION  XXXII—  THEOEEM. 

99.  The  area  of  a  spherical  triangle  is  equal  to  its  spherical  excess 
(the  right  angle  being  the  unit  of  angles  and  the  tri-rectangular 
triangle  the  unit  of  areas). 

For,   let  ABC  be  a  spherical  triangle.     Complete  the  great  circle 
ABA'B',  and  produce  the  arcs  AC  and  BC 
to  meet  this  circle  in  A'  and  B'. 

We  have,  by  the  figure, 

ABC  +  A  'BC  = 
ABC+  AB'C  =  \uueB, 

and  by  (93) 

C. 


The  sum  of  the  first  members  of  these  equations  is  equal  to  twice 
the  triangle  ABC,  plus  the  four  triangles  ABC,  A'BC,  AB'C, 
A'B'C,  which  compose  the  surface  of  the  hemisphere.  With  the 
system  of  units  adopted,  the  surface  of  the  hemisphere  is  expressed 
by  4;  therefore,  denoting  the  area  of  the  triangle  ABC  by  K,  and 
the  numerical  measures  of  its  angles  by  A,  B  and  C,  we  have  (97), 


whence 

K  =  A  +  B  +  C  —  2  =  spherical  excess. 

100.  Corollary.  The  same  reasoning,  in  connection  with  (94)  and 
(98),  may  be  employed  to  prove  that,  if  V  is  the  volume  of  a  spheri 
cal  triangular  pyramid  whose  base  is  the  spherical  triangle  ABC, 
and  if  the  unit  of  volume  is  the  volume  of  the  tri-rectangular  spheri 
cal  pyramid,  we  shall  have 

V=A  +  B+  C—2. 


268  GEOMETRY. 

101.  Scholium.  It  must  not  be  forgotten  that  the  preceding  results 
are  merely  the  expression  of  the  ratios  of  the  figures  considered  to 
the  adopted  units.  For  example,  suppose  the  angles  of  a  spherical 
triangle  are  given  in  degrees  as  follows  :  A  =  80°,  B  =  100°, 
C  =  120°  ;  then,  reducing  them  to  the  right  angle  as  the  unit, 


j=  _       4 

90         90         90  3 

therefore,  the  area  of  this  triangle  is  f  of  the  area  of  the  tri-rectangu- 
lar  triangle. 

Also,  the  volume  of  the  spherical  pyramid  of  which  this  triangle 
is  the  base  is  f  of  the  volume  of  the  tri-rectangular  spherical 
pyramid. 

Hence,  also,  it  follows  that  the  volumes  of  two  triangular  spherical 
pyramids  are  to  each  other  as  the  areas  of  their  bases. 


PROPOSITION  XXXIII.— THEOREM. 

102.  The  area  of  a  spherical  polygon  is  measured  by  the  sum  of  its 
angles  minus  the  product  of  two  right  angles  multiplied  by  the  number 
of  sides  of  the  polygon  less  two. 

Let  ABODE  be  a  spherical  polygon.  From 
any  vertex,  as  A,  draw  the  diagonals  A  C,  AD ; 
the  polygon  will  be  divided  into  as  many  tri 
angles  as  there  are  sides  less  two.  The  surface 
of  each  triangle  is  measured  by  the  sum  of  its 
angles  minus  two  right  angles ;  and  the  sum  of  all  the  angles  of  the 
triangles  is  equal  to  the  sum  of  the  angles  of  the  polygon ;  therefore 
the  surface  of  the  polygon  is  measured  by  the  sum  of  its  angles 
minus  two  right  angles  multiplied  by  the  number  of  triangles,  that 
is,  by  the  number  of  sides  of  the  polygon  less  two. 

103.  Corollary  I.  Denoting  the  number  of  sides  of  the  polygon 
by  n,  the  sum  of  its  angles  by  S,  and  its  area  by  K,  then,  with  the 
adopted  system  of  units,  we  have 

K=S—2(n  —  2)  =  £—  2?i  +  4. 

104.  Corollary  II.  The  tri-rectangular  pyramid  being  taken  as  the 
unit  of  volume,  the  volume  of  any  spherical  pyramid  will  have  the 


BOOK  viii.  269 

same  numerical  expression  as  the  area  of  its  base ;  that  is,  the  volume 
of  a  spherical  pyramid  is  to  the  volume  of  the  tri-reetangular  pyramid 
as  the  base  of  the  pyramid  is  to  the  tri-rectangular  triangle. 

Now  the  volume  of  the  tri-rectangular  pyramid  is  one-eighth  of  the 
volume  of  the  sphere,  and  the  tri-rectangular  triangle  is  one-eighth  of 
the  surface  of  the  sphere ;  therefore,  the  volume  of  a  spherical  pyramid 
is  to  the  volume  of  the  sphere  as  its  base  is  to  the  surface  of  the  sphere. 


SHOETEST    LINE    ON    THE    SUKFACE    OF   A    SPHEKE 
BETWEEN    TWO    POINTS. 

PKOPOSITION  XXXIV.— THEOREM. 

105.  The  shortest  line  that  can  be  drawn  on  the  surface  of  a  sphere 
between  two  points  is  the  arc  of  a  great  circle,  less  than  a  semi-circumfer 
ence,  joining  the  two  points. 

Let  AB  be  an  arc  of  a  great  circle,  less  than 
a  semi-circumference,  joining  any  two  points  A 
and  B  of  the  surface  of  a  sphere ;  and  let  C 
be  any  arbitrary  point  taken  in  that  arc.  Then 
we  say  that  the  shortest  line  from  A  to  B,  on 
the  surface  of  the  sphere,  must  pass  through  C. 

From  A  and  B  as  poles,  with  the  polar  dis 
tances  A  C  and  BC,  describe  circumferences  on  the  surface;  these 
circumferences  touch  at  C  and  lie  wholly  without  each  other.  For, 
let  M  be  any  point  in  the  circumference  whose  pole  is  A,  and  draw 
the  arcs  of  great  circles  AM,  BM,  forming  the  spherical  triangle 
AMB.  We  have,  by  (82),  AM  -f  BM  >  AB,  and  subtracting  from 
the  two  members  of  this  inequality  the  equal  arcs  AM  and  A  C,  we 
have  BM  >  BC',  therefore,  M  lies  without  the  circumference  whose 
pole  is  B. 

Now  let  AFQB  be  any  line  from  A  to  B,  on  the  surface  of  the 
sphere,  which  does  not  pass  through  the  point  C,  and  which  therefore 
cuts  the  two  circumferences  in  different  points,  one  in  F,  the  other  in 
O.  Whatever  may  be  the  nature  of  the  line  AF,  an  equal  line  can 
be  drawn  from  A  to  C;  for,  if  A  C  and  AF  be  conceived  to  be  drawn 
on  two  equal  spheres  having  a  common  diameter  passing  through  A, 
and  therefore  having  their  surfaces  in  coincidence,  and  if  one  of 

23* 


270  GEOMETRY. 

these  spheres  be  turned  upon  the  common  diameter  as  an  axis,  the 
point  A  will  be  fixed  and  the  point  F  will  come  into  coincidence  with 
C',  the  surfaces  of  the  two  spheres  continuing  to  coincide,  the  line 
AF  will  then  lie  on  the  common  surface  between  A  and  G.  For  the 
same  reason,  a  line  can  be  drawn  from  B  to  C,  equal  to  BG.  There 
fore,  a  line  can  be  drawn  from  A  to  B,  through  (7,  equal  to  the  sum 
of  AF  and  BG,  and  consequently  less  than  any  line  AFGB  that  does 
not  pass  through  C.  The  shortest  line  from  A  to  B  therefore  passes 
through  C,  that  is,  through  any,  or  every,  point  in  AB ;  consequently 
it  must  be  the  arc  AB  itself. 


"b 


fi° 


BOOK   IX. 


MEASUREMENT  OF  THE  THREE  ROUND  BODIES. 
THE  CYLINDER 

1.  DEFINITION.  The  area  of  the  convex,  or  lateral,  surface  of  a 
cylinder  is  called  its  lateral  area. 

2.  Definition.  A  prism  is  inscribed  in  a 
cylinder  when  its  bases  are  inscribed  in  the 
bases  of  the  cylinder. 

If  a  polygon  ABCDEF  is  inscribed  in 
the  base  of  a  cylinder,  planes  passed 
through  the  sides  of  the  polygon,  parallel 
to  the  elements  of  the  cylinder,  intersect 
the  cylinder  in  parallelograms,  ABB 'A', 
etc.  (VIII.  6),  which  evidently  determine 
a  prism  inscribed  in  the  cylinder. 

3.  Definition.  A  prism  is  circumscribed  about  a  cylinder  when  its 
bases  are  circumscribed  about  the  bases  of  the  cylinder. 

If  a  polygon  ABCD  is  circumscribed 
about  the  base  of  a  cylinder,  planes 
passed  through  the  sides  of  the  polygon, 
parallel  to  the  elements  of  the  cylinder, 
will  evidently  contain  the  elements,  <m', 
bbf,  etc.,  drawn  at  the  points  of  contact, 
and  be  tangent  to  the  cylinder  in  these 
elements.  The  intersection  of  these 
planes  with  the  plane  of  the  upper  base 
of  the  cylinder  will  therefore  determine 
a  polygon  A'B'C'D',  equal  to  ABCD, 

circumscribed  about  the  upper  base,  and  a  prism  will  be  formed  which 
is  circumscribed  about  the  cylinder. 

271 


D> 


272 


GEOMETRY. 


4.  Definition.  A  right  section  of  a  cylin 
der  is  a  section  made  by  a  plane  perpen 
dicular  to  its  elements ;  as  abcdef. 

The  intersection  of  the  same  plane  with 
an  inscribed  or  circumscribed  prism  is  a 
right  section  of  the  prism. 


5.  Definition.  Similar  cylinders  of  revolution  are  those  which 
are  generated  by  similar  rectangles  revolving  about  homologous 
sides. 


PROPOSITION  I.— THEOREM. 

6.  A  cylinder  is  the  limit  of  the  inscribed  and  circumscribed  prismst 
the  number  of  whose  faces  is  indefinitely  increased. 

Let  any  polygon  abed  be  inscribed  in  the  base  of  the  cylinder  ac'y 
and  at  the  vertices  of  this  polygon  let 
tangents  be  drawn  to  the  base  of  the 
cylinder  forming  the  circumscribed  poly 
gon  ABCD.  Upon  these  polygons  as 
bases  let  prisms  be  formed,  inscribed  in, 
and  circumscribed  about,  the  cylinder. 
We  shall  assume,  as  evident,  that  the 
convex  surface  of  the  cylinder  is  greater 
than  that  of  the  inscribed  prism  and 
less  than  that  of  the  circumscribed 
prism.* 

Suppose  the  arcs  ab,  be,  etc.,  to  be  bisected  and  polygons  to  be 
formed  having  double  the  number  of  sides  of  the  first ;  and  upon 
these  as  bases  suppose  prisms  to  be  constructed,  inscribed  and  circum 
scribed,  as  before;  and  let  this  process  be  repeated  an  indefinite 
number  of  times.  The  difference  between  the  convex  surface  of  the 
inscribed  prism  and  that  of  the  corresponding  circumscribed  prism 
will  continually  diminish  and  approach  to  zero  as  its  limit.  There- 

*  A  proof,  however,  can  be  given  analogous  to  that  of  (V.  32). 


BOOK   ix.  273 

fore  these  convex  surfaces  themselves  approach  to  the  convex  surface 
of  the  cylinder  as  their  common  limit. 

At  the  same  time,  it  is  evident  that  the  volumes  of  the  inscribed 
and  circumscribed  prisms  approach  to  the  volume  of  the  cylinder  as 
their  common  limit. 

7.  Scholium.  In  the  preceding  demonstration,  the  base  of  the  cylin 
der  is  not  required  to  be  a  circle,  but  may  be  any  closed  convex 
curve.  We  have,  however,  tacitly  assumed  that  the  curve  is  the 
limit  of  the  perimeters  of  the  inscribed  and  circumscribed  polygons ; 
a  principle  which  was  rigorously  proved  in  the  case  of  regular  poly 
gons  inscribed  in  a  circle. 


PEOPOSITION  II.— THEOKEM. 

8.  The  lateral  area  of  a  cylinder  is  equal  to  the  product  of  the 
perimeter  of  a  right  section  of  the  cylinder  by  an  element  of  the 
surface. 

Let  ABCDEF  be  the  base  and  A  A'  any 
element  of  a  cylinder,  and  let  the  curve 
abcdef  be  any  right  section  of  the  surface. 
Denote  the  perimeter  of  the  right  section 
by  P,  the  element  A  A '  by  E,  and  the  lat 
eral  area  of  the  cylinder  by  S. 

Inscribe  in  the  cylinder  a  prism 
ABCDEF  A'  of  any  number  of  faces. 
The  right  section,  abcdef,  of  this  prism  will 
be  a  polygon  inscribed  in  the  right  section 
of  the  cylinder  formed  by  the  same  plane. 

Denote  the  lateral  area  of  the  prism  by  s,  and  the  perimeter  of  its 
right  section  by  p ;  then,  the  lateral  edge  of  the  prism  being  equal 
to  E,  we  have  (VII.  16), 

s=p  XE. 

Let  the  number  of  lateral  faces  of  the  prism  be  indefinitely  increased, 
as  in  the  preceding  proposition ;  then  s  approaches  indefinitely  to  /S 
as  its  limit,  and  p  approaches  to  P;  therefore,  at  the  limit,  we  have 
(V.  31), 

S=PX  E. 
s 


274 


GEOMETRY. 


9.  Corollary  I.  The  lateral  area  of  a  right  cylinder  is  equal  to  the 
product  of  the  perimeter  of  its  base  by  its  altitude. 

10.  Corollary  II.  Let  a  cylinder  of  revolution  be 
generated  by  the  rectangle  whose  sides  are  R  and  H 
revolving  about  the  side  H.     Then,  R  is  the  radius  of 
the  base,  and  H  is  the  altitude  of  the  cylinder.     The 
perimeter  of  the  base  is  2xR  (V.  40),  and  hence,  for 
the  lateral  area  S  we  have  the  expression 

The  area  of  each  base  is  TrR*  (V.  43) ;  hence  the  total  area  T  of 
the  cylinder  of  revolution,  is  expressed  by 


T= 


11.  Corollary  III.  Let  8  and  s  de 
note  the  lateral  areas  of  two  similar 
cylinders  of  revolution  (4) ;  T  and  t 
their  total  areas;  R  and  r  the  radii 
of  their  bases ;  H  and  h  their  alti 
tudes.  The  generating  rectangles  be 
ing  similar,  we  have  (III.  12) 


R). 


therefore, 


H      R      H  +  R 

h        r         h  -f-  r 

2-RH_R    H^IP 

r      h         A2 

Jf2)       R    H+R 


IT 

h* 


That  is,  the  lateral  areas,  or  the  total  areas,  of  similar  cylinders  of  revo 
lution  are  to  each  other  as  the  squares  of  their  altitudes,  or  as  the  squares 
of  the  radii  of  their 


BOOK     IX. 


275 


PEOPOSITION  III.—  PKOBLEM. 

12.  The  volume  of  a  cylinder  is  equal  to  the  product  of  its  base  by  its 
altitude. 

Let  the  volume  of  the  cylinder  be  denoted 
by  F,  its  base  by  J5,  and  its  altitude  by  H. 
Let  the  volume  of  an  inscribed  prism  be  de 
noted  by  F',  and  its  base  by  B'  \  its  altitude 
will  also  be  Ht  and  we  shall  have  (VII.  38) 

V  =  B'  X  H. 

Let  the  number  of  faces  of  the  prism  be 
indefinitely  increased,  as  in  (8)  ;  then  the  limit 
of  V  is  F,  and  the  limit  of  B'  is  B\  therefore  (V.  31), 


13.  Corollary  I.  Let  F  be  the  volume  of  a  cylinder  of  revolution, 
R  the  radius  of  its  base,  and  H  its  altitude  ;  then  the  area  of  its 
base  is  xR2  (V.  43)  ;  and  therefore 


14.  Corollary  II.  Let  F  and  v  be  the  volumes  of  two  similar  cyl 
inders  of  revolution  ;  R  and  r  the  radii  of  their  bases  ;  H  and  h 
their  altitudes;  then,  the  generating  rectangles  being  similar,  we 
have 

H_R 
h  ~~  r 
and 


F 
v 


w  JT^-HJ 

r2  '  h=~  h3 


R 


that  is,  the  volumes  of  similar  cylinders  of  revolution  are  to  each  other 
as  the  cubes  of  their  altitudes,  or  as  the  cubes  of  their  radii. 


276 


GEOMETRY. 


THE    CONE. 

15.  Definition.  The  area  of  the  convex,  or  lateral,  surface  of  a  cone 
is  called  its  lateral  area. 

16.  Definition.  A  pyramid  is  inscribed  in  a 
cone  when  its  base  is  inscribed  in  the  base  of 
the  cone,  and  its  vertex  coincides  with  the 
vertex  of  the  cone. 

If  a  polygon  AB  CD  is  inscribed  in  the  base 
of  a  cone  and  planes  are  passed  through  its 
sides  and  the  vertex  8  of  the  cone,  these 
planes  intersect  the  convex  surface  of  the 
cone  in  right  lines  (VIII.  18)  and  determine 
a  pyramid  inscribed  in  the  cone. 

17.  Definition.   A    pyramid    is    circum 
scribed  about  a  cone  when  its  base  is  cir 
cumscribed  about  the  base  of  the  cone,  and 
its  vertex  coincides  with  the  vertex  of  the 
cone. 

If  a  polygon  ABCD  is  circumscribed 
about  the  base  of  a  cone,  its  points  of  con 
tact  with  the  base  being  a,  b,  c,  d,  and 
planes  are  passed  through  its  sides  and  the 
vertex  S  of  the  cone,  these  planes  will  be  tangent  to  the  cone  in 
the  elements  Sa,  Sb,  etc.  (VIII.  21),  and  will  determine  a  pyramid 
circumscribed  about  the  cone. 

18.  Definition.  A  truncated  cone  is  the  portion  of  a  cone  included 
between  its  base  and  a  plane  cutting  its  convex  surface. 

When  the  cutting  plane  is  parallel  to 
the  base,  the  truncated  cone  is  called  a 
frustum  of  a  cone;  as  AB  CD-abed.  The 
altitude  of  a  frustum  is  the  perpendicular 
distance  Tt  between  its  bases. 

If  a  pyramid  is  inscribed  in  the  cone, 
the  cutting  plane  determines  a  truncated 
pyramid  inscribed  in  the  truncated  cone ; 
and  if  a  pyramid  is  circumscribed  about 


BOOK    IX. 


277 


the  cone,  the  cutting  plane  determines  a  truncated  pyramid  circum 
scribed  about  the  truncated  cone. 

19.  Definition.  In   a   cone   of  revolution,   as 
S-ABC,   generated   by   the  .revolution   of  the 
right  triangle  SA  0  about  the  axis  SO,  all  the 
elements,  SA,  SB,  etc.,  are  equal ;  and  any  ele 
ment  is  called  the  slant  height  of  the  cone. 

In  a  cone  of  revolution,  the  portion  of  an  ele 
ment  included  between  the  parallel  bases  of  a 
frustum,  as  Aa,  or  Bb,  is  called  the  slant  height 
of  the  frustum. 

20.  Definition.  Similar  cones  of  revolution  are  those   which   are 
generated  by  similar  right  triangles   revolving  about  homologous 
sides. 

PEOPOSITION  IV.— THEOKEM. 

21.  A  cone  is  the  limit  of  the  inscribed  and  circumscribed  pyramids, 
the  number  of  whose  faces  is  indefinitely  increased. 

The  demonstration  is  precisely  the  same  as  that  of  Proposition  I., 
substituting  a  cone  for  a  cylinder,  and  pyramids  for  prisms. 

22.  Corollary.  A  frustum  of  a  cone  is  the  limit  of  the  frustums  of 
the  inscribed  and  circumscribed  frustums  of  pyramids,  the  number 
of  whose  faces  is  indefinitely  increased. 


PEOPOSITION  V.— THEOKEM. 

23.   The  lateral  area  of  a  cone  of  revolution  is  equal  to  the  product 
of  the  circumference  of  its  base  by  half  its  slant  height. 

Let  S-MNPQ  be  a  cone  generated  by  the 
revolution  of  the  right  triangle  SOM  about 
the  axis  SO.  Denote  its  lateral  area  by  S, 
the  circumference  of  its  base  by  (7,  and  its 
slant  height  SMloy  L. 

Circumscribe  about  the  base  any  regular 

polygon  ABCD,  and  upon  this  polygon  as  a 

base  construct  a  regular  pyramid  S-ABCD 

circumscribed  about  the  cone.     Denote  the 

24 


278 


GEOMETRY. 


lateral  area  of  the  pyramid  by  s,  and  the  perimeter  of  its  base  oy 
p  ;  its  slant  height  is  the  same  as  that  of  the  cone,  since  it  is  an  ele 
ment  of  contact,  as  SM  or  SN\  therefore,  we  have  (VII.  47), 

L 

s=PX~. 

The  number  of  lateral  faces  of  the  pyramid  being  indefinitely  in 
creased,  s  approaches  indefinitely  to  S,  and  p  approaches  indefinitely 
to  C;  therefore,  at  the  limit,  we  have  (V.  31), 


24.    Corollary  I.  If  R  is  the  radius  of  the  base,  we  have  C  = 
(V.  40)  ;  hence 


S  =  2*R  X  -  = 


The  area  of  the  base  being  ?rjR2,  the  total  area  T  of  the  cone  is 
T  =  nRL  +  TrE2  =  *R  (L  +  12). 

25.  Corollary  II.  Hence,  by  the  same  process  as  was  employed  in 
(11),  we  can  prove  that  the  lateral  areas,  or  the  total  areas,  of  similar 
cones  of  revolution  are  to  each  other  as  the  squares  of  their  slant  heights, 
or  as  the  squares  of  their  altitudes,  or  as  the  squares  of  the  radii  of 
their  bases. 

PROPOSITION  VI.—  THEOREM. 

26.  The  lateral  area  of  a  frustum  of  a  cone  of  revolution  is  equal  to 
the  half  sum  of  the  circumferences  of  its  bases  multiplied  by  its  slant 
height. 

The  plane  which  cuts  off  the  frustum 
MNPm,  from  the  cone  S-MNP,  also  cuts 
off  from  any  circumscribed  pyramid  a 
frustum,  as  ABCDa,  the  lateral  area  of 
.which  is  equal  to  the  half  sum  of  the  pe 
rimeters  of  its  bases  multiplied  by  its  slant 
height  Mm  (VII.  48).  When  the  number 
of  faces  of  the  frustum  of  the  pyramid  is 
indefinitely  increased,  its  lateral  area  ap 
proaches  indefinitely  to  that  of  the  frustum 


BOOK     IX. 


279 


of  the  cone,  and  the  perimeters  of  its  bases  approach  indefinitely  to 
the  circumferences  of  the  bases  of  the  frustum  of  the  cone ;  and  the 
slant  height  Mm  is  common.  Hence,  if  we  express  by  area  Mm, 
the  area  of  the  surface  generated  by  the  revolution  of  Mm  about 
the  axis,  which  is  the  lateral  area  of  the  frustum  of  the  cone ;  and 
by  circ.  OM,  and  circ.  om,  the  circumferences  of  the  bases  whose  radii 
are  OM  and  om  ;  we  shall  have,  at  the  limit, 

area  Mm  =  -|  (circ.  OM  -f  circ.  om)  X  Mm. 


27.  Corollary.  Let  IK  be  the  radius  of  a 
section  of  the  frustum  equidistant  from  its 
bases;  then,  IZT=  1(0^+  om),  (1. 124),  and 
since  circumferences  are  proportional  to  their 
radii,  circ.  IK  =  -|-  (circ.  OM  -{-  circ.  om)  ; 

therefore, 

area  Mm  =  circ.  IK  X  Mm ; 


that  is,  the  lateral  area  of  a  frustum  of  a  cone  of  revolution  is  equal 
to  the  circumference  of  a  section  equidistant  from  its  bases  multiplied  by 
its  slant  height. 


PKOPOSITION  VII.—  THEOKEM. 

28.  The  volume  of  any  cone  is  equal  to  one-third  of  the  product  of 
its  base  by  its  altitude. 

Let  the  volume  of  the  cone  be  denoted  by 
V,  its  base  by  B,  and  its  altitude  by  H. 

Let  the  volume  of  an  inscribed  pyramid  be 
denoted  by  F',  and  its  base  by  B'  ',  its  alti 
tude  will  also  be  H,  and  we  shall  have 
(VII.  54), 

V  =  i  B'  X  H. 

When  the  number  of  lateral  faces  of  the 

pyramid  is  indefinitely  increased,   V'  approaches  indefinitely  to  F, 
and  -B'  to  -B;  therefore,  at  the  limit,  we  have 


280  GEOMETRY. 

29.   Corollary  I.  If  the  cone  is  a  cone  of  revolution,  let  E  be  the 
radius  of  the  base,  then  B  =  irR  2,  and  we  have 


30.  Corollary  II.  Let  R  and  r  be  the  radii  of  the  bases  of  two 
similar  cones  of  revolution  ;  H  and  h  their  altitudes,  V  and  v  their 
volumes  ;  then,  the  generating  triangles  being  similar,  we  have 


and  hence 


that  is,  similar  cones  of  revolution  are  to  each  other  as  the  cubes  of  their 
altitudes,  or  as  the  cubes  of  the  radii  of  their  bases. 


PKOPOSITION  VIII.—  THEOREM. 

31.  A  frustum  of  any  cone  is  equivalent  to  the  sum  of  three  cones 
whose  common  altitude  is  the  altitude  of  the  frustum,  and  whose  bases 
are  the  lower  base,  the  upper  base,  and  a  mean  proportional  between 
the  bases  of  the  frustum. 

Let  V  denote  the  volume  of  the  frustum, 
B  its  lower  base,  b  its  upper  base,  and  h  its 
altitude. 

Let  V  denote  the  volume  of  an  inscribed 
frustum  of  a  pyramid,  B'  its  lower  base,  and  b' 
its   upper   base;   its   altitude  will   also   be   h,  and  we  shall   have 
(VII.  59), 


When  the  number  of  lateral  faces  of  the  frustum  of  a  pyramid  is 
indefinitely  increased,  V,  B'  and  b'f,  approach  indefinitely  to  V,  B 
and  b,  respectively  ;  therefore,  at  the  limit,  we  have 


which  is  the  algebraic  expression  of  the  theorem. 


BOOK      IX.  281 

32.   Corollary.  If  the  frustum  is  that  of  a  cone  of  revolution,  and 
the  radii  of  its  bases  are  R  and  r,  we  shall  have 


and  consequently, 

V= 


THE  SPHEKE. 

33.  Definition.  A  spherical  segment  is  a  portion  of  a  sphere  in 
cluded  between  two  parallel  planes. 

The  sections  of  the  sphere  made  by  the  parallel  planes  are  the 
bases  of  the  segment  ;  the  distance  of  the  planes  is  the  altitude  of  the 
segment. 

Let  the  sphere  be  generated  by  the  revolution  of  _£  _  f 
the  semicircle  EBF  about  the  axis  EF-,  and  let  Aa 
and  Bb  be  two  parallels,  perpendicular  to  the  axis. 
The  solid  generated  by  the  figure  ABba  is  a  spheri 
cal  segment  ;  the  circles  generated  by  Aa  and  Bb  are 
its  bases  ;  and  ab  is  its  altitude. 

If  two  parallels  Aa  and    TE  are  taken,  one  of 
which  is  a  tangent  at  E,  the  solid  generated  by  the 
figure  EAa  is  a  spherical  segment  having  but  one 
base,  which  is  the  section  generated  by  Aa.     The  segment  is  still  in 
cluded  between  two  parallel  planes,  one  of  which  is  the  tangent 
plane  at  E,  generated  by  the  line  ET. 

34.  Definition.  A  zone  is  a  portion  of  the  surface  of  a  sphere  in 
cluded  between  two  parallel  planes. 

The  circumferences  of  the  sections  of  the  sphere  made  by  the 
parallel  planes  are  the  bases  of  the  zone  ;  the  distance  of  the  planes 
is  its  altitude. 

A  zone  is  the  curved  surface  of  a  spherical  segment. 

In  the  revolution  of  the  semicircle  EBF  about  EF,  an  arc  AB 
generates  a  zone  ;  the  points  A  and  B  generate  the  bases  of  the  zone  ; 
and  the  altitude  of  the  zone  is  ab. 

An  arc,  EA,  one  extremity  of  which  is  in  the  axis,  generates  a 

24* 


282  GEOMETRY. 

zone  of  one  base,  which  is  the  circumference  described  by  the  ex 
tremity  A. 

35.  Definition.  When  a  semicircle  revolves  about  its  diameter,  the 
solid  generated  by  any  sector  of  the  semicircle  is  called  a  spherical 
sector.  . 

Thus,  when  the  semicircle  EBF  revolves  about  EF,  the  circular 
sector  COD  generates  a  spherical  sector. 

The  spherical  sector  is  bounded  by  three  curved  surfaces ;  namely, 
the  two  conical  surfaces  generated  by  the  radii  OC  and  OD,  and  the 
zone  generated  by  the  arc  CD.  This  zone  is  called  the  base  of  the 
spherical  sector. 


PROPOSITION  IX.— LEMMA. 

36.  The  area  of  the  surface  generated  by  a  straight  line  revolving 
about  an  axis  in  its  plane,  is  equal  to  the  projection  of  the  line  on  the 
axis  multiplied  by  the  circumference  of  the  circle  whose  radius  is  the 
perpendicular  erected  at  the  middle  of  the  line  and  terminated  by  the 
axis, 

Let  AB  be  the  straight  line  revolving  about 
the  axis  XY;  ab  its  projection  on  the  axis ;  01 
the  perpendicular  to  it,  at  its  middle  point  7, 
terminating  in  the  axis ;  then, 

area  AB  =  ab  X  circ.  01. 


For,  draw  IK  perpendicular,  and  AH  par 
allel  to  the  axis.  The  area  generated  by  AB  is 
that  of  a  frustum  of  a  cone  ;  hence  (27), 

area  AB  =  AB  X  circ.  IK 

Now  the  triangles  ABH  and  IOK,  having  their  sides  perpendicular 
each  to  each,  are  similar  (III.  33),  hence 


=  IK:  01, 

or,  since  circumferences  are  proportional  to  their  radii, 

ab  :  AB  =  circ.  IK  :  circ.  01, 
whence 


therefore, 


BOOK     IX. 
AB  X  circ.  IK  =  ab  X  circ.  01, 

area  AB  =  ab  X  circ.  01. 


If  AB  is  taken  parallel  to  the  axis,  the  result  is 
the  same,  and  in  fact  has  already  been  proved,  since 
in  this  case  the  surface  generated  is  that  of  a  cylin 
der  whose  radius  is  01  and  whose  altitude  is  ab  (9). 


283 


PKOPOSITION  X.— THEOREM. 

37.  The  area  of  a  zone  is  equal  to  the  product  of  its  altitude  by  the 
circumference  of  a  great  circle. 

Let  the  sphere  be  generated  by  the  revolution  of 
the  semicircle  EBF  about  the  axis  EF;  and  let  the 
arc  AD  generate  the  zone  whose  area  is  required. 

Let  the  arc  AD  be  divided  into  any  number  of 
equal  parts,  AS,  BC,  CD.  The  chords  AB,  BC, 
CD,  form  a  regular  broken  line,  which  differs  from  a 
portion  of  a  regular  polygon  only  in  this,  that  the 
arc  subtended  by  one  of  its  sides,  as  AB,  is  not 
necessarily  an  aliquot  part  of  the  whole  circumfer 
ence.  The  sides  being  equidistant  from  the  centre,  a  circle  described 
with  the  perpendicular  OI,  let  fall  from  the  centre  upon  any  side, 
would  touch  all  the  sides  and  be  inscribed  in  the  regular  broken  line. 
Drawing  the  perpendiculars  Aa,  Bb,  Cc,  Dd,  we  have  by  the  preced 
ing  LEMMA, 

area  AB  =  ab  X  circ.  01, 
area  BC  =  be  X  circ.  01, 
area  CD  =  cd  X  circ.  01, 

the  sum  of  which  is 


or 


area  ABCD  =  (ab  +  be  +  cd)  X  circ.  01, 


area  ABCD  =  ad  X  circ.  01. 


This  being  true  whatever  the  number  of  sides  of  the  regular  broken 
line,  let  that  number  be  indefinitely  increased ;  then  area  ABCD, 


284  GEOMETRY. 

generated  by  the  broken  line,  approaches  indefinitely  to  the  area  of 
the  zone  generated  by  the  arc  AD,  and  cire.  01  approaches  indefi 
nitely  to  cire.  OE,  or  the  circumference  of  a  great  circle  ;  hence,  at 
the  limit,  we  have 

area  of  zone  AD  =  ad  X  cire.  OE, 

which  establishes  the  theorem. 

38.   Corollary  I.  Let  S  denote  the  surface  of  the  zone  whose  alti 
tude  is  Hj  the  radius  of  the  sphere  being  R  ;  then, 

S  = 


39.  Corollary  II.  Zones  on  the  same  sphere,  or  on  equal  spheres, 
are  to  each  other  as  their  altitudes. 

40.  Corollary  III.  Let  the  arc  AD  generate  a  zone 

of  a  single  base ;  its  area  is  D/ 

AdX2x.OA  =  x.AdxAB  =  *.  AD2  (III.  47); 

T  O 

that  is,  a  zone  of  one  base  is  equivalent  to  the  circle 
whose  radius  is  the  chord  of  the  generating  arc  of  the 
zone. 

PROPOSITION  XI.— THEOREM. 

41.  The  area  of  the  surface  of  a  sphere  is  equal  to  the  product  of  its 
diameter  by  the  circumference  of  a  great  circle. 

This  follows  directly  from  the  preceding  proposition,  since  the  sur 
face  of  the  whole  sphere  may  be  regarded  as  a  zone  whose  altitude  is 
the  diameter  of  the  sphere. 

42.  Corollary  I.  Let  S  denote  the  area  of  the  surface  of  a  sphere 
whose  radius  is  R ;  then 


that  is,  the  surface  of  a  sphere  is  equivalent  to  four  great  circles. 

43.   Corollary  II.  Let  S  and  Sr  be  the  surfaces  of  two  spheres 
whose  radii  are  R  and  R  '  ;  then, 


hence,  the  surfaces  of  two  spheres  are  to  each  other  as  the  squares  of 
their  diameters,  or  as  the  squares  of  their  radii. 


BOOK  ix.  285 


PKOPOSITION  XII.— LEMMA. 

44.  If  a  triangle  revolves  about  an  axis  situated  in  its  plane  and 
passing  through  the  vertex  without  crossing  its  surface,  the  volume 
generated  is  equal  to  the  area  generated  by  the  base  multiplied  by  one- 
third  of  the  altitude. 

Let  ABC  be  the  triangle  revolving  about  an  axis  XY  passing 
through  the  vertex  A ;  then,  the  volume  generated  is  equal  to  the 
area  generated  by  the  base  BC  multiplied  by  one-third  of  the  alti 
tude  AD. 

We  shall  distinguish  three  cases  : 

1st.  When  one  of  the  sides  of  the  triangle,  as  ABt  lies  in  the  axis. 
(Figs.  1  and  2.) 

Fig.  1.  Fig.  2. 

G 


Draw  CE  perpendicular  to  the  axis.  According  as  this  perpen 
dicular  falls  within  the  triangle  (Fig.  1)  or  without  it  (Fig.  2),  the 
volume  generated  is  the  sum  or  the  difference  of  the  cones  generated 
by  the  right  triangles  ACE  and  BCE.  The  volumes  of  these  cones 
are  (29), 

voLACE=$n.  CE2  X  AE, 

vol.  BCE  =  |  TT  .  CE*  x  BE-, 

if  we  take  their  sum,  we  have  in  Fig.  1,  AE  -\-  BE  =  AB;  if  we 
take  their  difference,  we  have  in  Fig.  2,  AE  —  BE  =  AB  ;  there 
fore,  in  either  case, 


=  ^x.  CE  X  CEXAB-, 

or,  since  CE  X  AB  is  double  the  area  of  the  triangle  which  is  also 
expressed  by  BC  X  AD  (IV.  13), 

voLABC=$*.  CEXBCX  AD. 

But  TT  .  CE  X  BC  is  the  measure  of  the  surface  generated  by  BC 
(24);  therefore, 

vol.  ABC  =  area  BC  X 


286  GEOMETRY. 

2d.  When  the  triangle  has  only  the 
vertex  A  in  the  axis,  and  the  base  B  C 
when  produced  meets  the  axis  in  F 
(Fig.  3).  *'  A  F  * 

The  volume  generated  is  then  the 

difference  of  the  volumes  generated  by  the  triangles  A  CF  and  ABFt 
and,  by  the  first  case,  these  volumes  are 

vol.  ACF  =  area  FC  X  4  A  D, 
vol.  ABF  =  area  FB  X 


the  difference  of  which  is 
vol.  ABC  =  (area  FC—  area  FB}  X  $AD  =  area  BC  X 

3d.  When  the  triangle  has  only  the  vertex  A  in  the  axis,  and  the 
base  BCis  parallel  to  the  axis  (Figs.  4  and  5). 

Fig.  4.                                                            Fig.  5. 
CD  B  DC        B 


K     A 


II 


The  volume  generated  is  the  sum  (Fig.  4),  or  the  difference  (Fig. 
5),  of  the  volumes  generated  by  the  right  triangles  ABD  and  A  CD. 

Draw  BHaud  CK  perpendicular  to  the  axis.  The  volume  gener 
ated  by  the  triangle  ABD  is  the  difference  of  the  volumes  of  the 
cylinder  generated  by  the  rectangle  ADBH  and  the  cone  generated 
by  the  triangle  ABH  ;  therefore, 


or,  since  2*  .  A  D  X  BD  is  the  lateral  area  of  the  cylinder  gener 
ated  by  the  rectangle  AHBD  (9), 

vol.  ABD  =  area  BD^ 
and  in  the  same  manner  we  have 

vol.  ACD  =  area  CD  X 


BOOK   ix.  287 

Taking  the  sum  of  these  (Fig.  4),  or  their  difference  (Fig.  5),  we 

have 

vol.  ABC=  area  BC  X  \AD. 

Therefore,  in  all  cases,  the  volume  generated  by  the  triangle  is  equal 
to  the  area  generated  by  its  base  multiplied  by  one-third  of  its 
altitude. 


PROPOSITION  XIII.— THEOREM. 

45.  The  volume  of  a  spherical  sector  is  equal  to  the  area  of  the  zone 
which  forms  its  base  multiplied  by  one-third  the  radius  of  the  sphere. 

Let  the  sphere  be  generated  by  the  revolution  of 
the  semicircle  EBF  about  the  axis  EF ;  and  let  the 
circular  sector  AOD  generate  a  spherical  sector 
whose  volume  is  required. 

Inscribe  in  the  arc  AD  a  regular  broken  line 
ABCD,  as    in  Proposition  X.,  forming   with    the 
radii    OA    and    OD    a  regular    polygonal    sector 
OABCD.      Decompose  this  polygonal  sector  into 
triangles  A  OB,  BOG,  COD,  by  drawing  radii  to 
its  vertices.     Taking  the  sides  AB,  BC,  CD,  as  bases,  the  perpen 
dicular  01  from  the  centre  upon  any  side  is  the  common  altitude 
of  these  triangles. 

The  volume  generated  by  the  polygonal  sector  is  the  sum  of  the 
volumes  generated  by  the  triangles,  and  the  volume  generated  by 
any  triangle  is  equal  to  the  area  of  its  base  multiplied  by  one-third 
of  its  altitude  01  (44) ;  therefore, 

vol.  OABCD  =  area  ABCD  X    —  • 

o 

When  the  number  of  sides  of  the  regular  polygonal  sector  is  in 
definitely  increased,  vol.  OABCD  approaches  indefinitely  to  the 
volume  of  the  spherical  sector  OAD,  area  ABCD  to  the  area  of  the 
zone  AD,  and  01  to  the  radius  OA  of  the  sphere ;  therefore,  at  the 
limit,  we  have 

vol.  spherical  sector  OAD  =  zone  A  D  X 
which  establishes  the  theorem. 


288  GEOMETRY. 


PROPOSITION  XIV.—  THEOREM. 

46.  The  volume  of  a  sphere  is  equal  to  the  area  of  its  surface  multi 
plied  by  one-third  of  its  radius. 

This  follows  directly  from  the  preceding  proposition  ;  for,  if  a  cir 
cular  sector  is  increased  until  it  becomes  the  semicircle  which  gener 
ates  the  sphere,  the  spherical  sector  which  it  generates  becomes  the 
sphere  itself,  and  its  surface  becomes  the  surface  of  the  sphere. 

47.  Corollary  I.  If  F  denotes  the  volume  of  a  sphere  whose  radius 
is  R,  we  have  (42) 


Or,  if  D  is  the  diameter  of  the  sphere,  whence  D3  =  (2JQ3  =  8R3, 


48.   Corollary  II.   The  volumes  of  two  spheres  are  to  each  other  as  the 
cubes  of  their  radii,  or  as  the  cubes  of  their  diameters. 


PROPOSITION  XV.— THEOREM. 

49.  The  solid  generated  by  a  circular  segment  revolving  about  a 
diameter  exterior  to  it,  is  equivalent  to  one-sixth  of  the  cylinder  whose 
radius  is  the  chord  of  the  segment  and  whose  altitude  is  the  projection 
of  that  chord  on  the  axis. 

Let  ANBIA  be  a  circular  segment  revolving 
about  the  diameter  EF,  and  ab  the  projection  of' 
the  chord  AB  on  the  axis.  The  volume  generated 
is  the  difference  of  the  volumes  generated  by  the 
circular  sector  A  OB  and  the  triangle  A  OB.  Draw 
ing  01  perpendicular  to  AB,  we  have  (45),  (44), 
(38)  and  (36), 

vol.  sph.  sector  A  OB  =  zone  AB  X  \OA  =  f  ~  .~OA2 
vol.  triangle  AOB  =  area  AB  X  £  01  =  f  TT  .  Of2 

the  difference  of  which  gives 

vol.  segment  ANB  =  %*(OA2  —  ~OT)  X  ab. 


BOOK  ix.  289 


But  -Qtf  _  OJ2  =  ZP  =  Jl£2;  hence 
vol.  segment  ANB  =  -J-w  . 

which  establishes  the  theorem,  since  TT  .  ZB  2  .  «6  is  the  volume  of  the 
cylinder  whose  radius  is  AB  and  whose  altitude  is  ab  (13). 

PKOPOSITION  XVL—  THEOREM. 

50.  The  volume  of  a  spherical  segment  is  equal  to  the  half  sum  of  its 
bases  multiplied  by  its  altitude  plus  the  volume  of  a  sphere  of  which 
that  altitude  is  the  diameter. 

Let  Aa  and  Bb  be  the  radii  of  the  bases  of  a 
spherical  segment,  and  ab  its  altitude,  so  that  the 
segment  is  generated  by  the  revolution  of  the  figure    B  ^ 
ANBba  about  the  axis  EF. 

The  segment  is  the  sum  of  the  solid  generated  by 

the  circular  segment  ANB  and  the  frustum  of  a  cone 

generated  by  the  trapezoid  ABba  ;  hence,  denoting 

the  volume  of  the  spherical  segment  by  F,  we  have 

•  C49)  and  (32), 

V=%*.AB2.cib  +  ^7t.(Bb2  +  Aa2  +  Bb  .  Aa)  .  ab. 

Drawing  AH  parallel  to  JEF,  we  have  BH  =  Bb  —  Aa,  and 
hence 

BH*='Bb*  +  Aa*  —  2Bb  .  Aa, 
and 


=  AH*-+  BH*  =  ab*  +  Bb2  +  ~Aa   —  2Bb  .  Aa. 
Substituting  this  value  of  AB  2,  we  have,  after  reduction, 

F=  K*  •  ^2  +  *  -~Ia^  -  ab  +  ^  -  °*3> 

which  establishes  the  theorem,  since  -R  .'Bb2  and  K  .  Aa2  represent  the 
bases  of  the  segment,  and  ITT  .~ab  3  is  the  volume  of  the  sphere  whose 
diameter  is  ab  (47). 

51.   Corollary.  Denoting  the  radii  of  the  bases  of  the  spherical 
segment  by  E  and  r,  and  its  altitude  by  h,  we  have,  for  its  volume, 


25 


290  GEOMETRY. 

If  the  point  A  coincides  with  E,  the  upper  base  becomes  zero,  and 
the  solid  generated  becomes  a  segment  of  one  base.  Therefore, 
making  r  =  0  in  the  above  expression,  the  volume  of  a  spherical 
segment  of  one  base  is 


PEOPOSITION  XVII.— THEOEEM. 

52.  The  volume  of  a  spherical  pyramid  is  equal  to  the  area  of  its  base 
multiplied  by  one-third  of  the  radius  of  the  sphere. 

For,  let  v  denote  the  volume  of  a  spherical  pyramid,  and  s  the 
area  of  the  spherical  polygon  which  forms  its  base.  Let  V,  S  and 
R  denote  the  volume,  surface  and  radius  of  the  sphere;  then 
(VIII.  104), 

V  8  V 

-  =  -,  whence  *  =  •  X  £• 


V 

But  -  =  %R  (46) ;  therefore, 


v  =  t  X 


APPENDIX    I. 


EXERCISES  IN  ELEMENTARY  GEOMETRY. 


EXERCISES  IN  ELEMENTARY  GEOMETRY. 


IN  order  to  make  these  exercises  progressive  as  to  difficulty,  and  to  bring 
them  fairly  within  the  grasp  of  the  student  at  the  successive  stages  of  his 
progress,  many  of  them  are  accompanied  by  diagrams  in  which  the  necessary 
auxiliary  lines  are  drawn,  or  by  references  to  the  articles  in  the  GEOMETRY 
on  which  the  exercise  immediately  depends,  or  by  both.  *These  aids  are  less 
and  less  freely  given  in  the  later  exercises,  and  the  student  is  finally  left 
wholly  to  his  own  resources. 


GEOMETRY.— BOOK  I. 

THEOEEMS. 

1.  The  sum  of  the  three  straight  lines  drawn  from  any  point 
within  a  triangle  to  the  three  vertices,  is  less  than  the  sum 
and  greater  than  the  half  sum  of  the  three  sides  of  the  tri 
angle  (I.  33,  66). 


2.  The  medial  line  to  any  side  of  a  triangle  is  less  than  the 
half  sum  of  the  other  two  sides,  and  greater  than  the  excess 
of  that  half  sum  above  half  the  third  side  (I.  66,  67,  112). 


3.  The  sum  of  the  three  medial  lines  of  a  triangle  is  less 
than  the  perimeter  (sum  of  the  three  sides),  and  greater  than 
the  semi-perimeter  of  the  triangle. 


4.  If  from  two  points,  A  and  j5,  on  the  same 
side  of  a  straight  line  MN,  straight  lines,  AP,  BP, 
are  drawn  to  a  point  P  in  that  line,  making  with  it 
equal  angles  APM  and  JBPN,  the  sum  of  the  lines 
APsmd  JSP  is  less  than  the  sum  of  any  other  two 
lines,  AQ  and  BQ,  drawn  from  A  and  B  to  any 
other  point  Q  in  MN(l.  83,  38,  66). 

25* 


294 


EXERCISES. 


5.  If  from  two  points,  A  and  B,  on  opposite  sides 
of  a  straight  line  MN,  straight  lines  AP,  BP,  are 
drawn  to  a  point  P  in  that  line,  making  with  it  equal 

angles  APN  and  BPN,  the  difference  of  the  lines  P\V 

AP  and  BP  is  greater  than  the  difference  of  any  B 

other  two  straight  lines  A  Q  and  B  Q,  drawn  from  A 
and  B  to  any  other  point  Q  in  MN. 

6.  The  three  straight  lines  joining  the  middle  points  of  the  sides  of  a  tri 
angle  divide  the  triangle  into  four  equal  triangles  (I.  122). 


Y.  The  straight  line  AE  which  bisects  the  angle  ex- 
terior  to  the  vertical  angle  of  an  isosceles  triangle  AB  C-, 
is  parallel  to  the  base  BC. 


8.  In  any  right  triangle,  the  straight  line  drawn  from 
the  vertex  of  the  right  angle  to  the  middle  of  the  hy 
potenuse  is  equal  to  one-half  the  hypotenuse  (I.  121, 

38,  46). 


9.  If  one  of  the  acute  angles  of  a  right  triangle  is  double  the  other,  the 
hypotenuse  is  double  the  shortest  side  (Ex.  8),  (I.  69,  86,  90). 

10.  If  ABC  is  any  right  triangle,  and  if  from 
the  acute  angle  JL,  AD  is  drawn  cutting  BCin  E 

and  a  parallel  to  AC  in  D  so  that  ED  =  2AB;  j 
then,  the  angle  D  A  C  is  one-third  the  angle  B  A  C.  I* 
(Ex.  8),  (L  69,  86,  49). 

11.  If  BC  is  the  base  of  an  isosceles  triangle  ABC,  and  BD  is  drawn 
perpendicular  to  AC,  the  angle  DBC  is  equal  to  one-half  the  angle  A. 
(L  73). 


12.  If  from  a  variable  point  in  the  base  of  an  isosceles  tri 
angle  parallels  to  the  sides  are  drawn,  a  parallelogram  is  formed 
whose  perimeter  is  constant. 

13.  If  from  a  variable  point  P  in  the  base  of  an  isos 
celes  triangle  AB  (7,  perpendiculars,  PM,  Pty  to  the  sides, 
are  drawn,  the  sum  of  PM  and  PJVis  constant,  and  equal 
to  the  perpendicular  from  C  upon  AB  (I.  104,  83). 


What  modification  of  this  statement  is  required  when  Pis  taken  in  BC 
produced  ? 


EXERCISES. 


295 


14.  If  from  any  point  within  an  equilateral  triangle,  per 
pendiculars  to  the  three  sides  are  drawn,  the  sum  of  these 
lines  is  constant,  and  equal  to  the  perpendicular  from  any 
vertex  upon  the  opposite  side  (Ex.  l£). 


What  modification  of  this  statement  is  required  when  the  point  is  taken 
without  the  triangle  ? 


15.  If  ABC  is  an  equilateral  triangle,  and  if  BD  and 
CD  bisect  the  angles  B  and  C,  the  lines  DE,  DP,  paral 
lel  to  AB,  AC,  respectively,  divide  EG  into  three  equal 
parts. 


16.  The  locus  of  all  the  points  which  are  equally  distant  from  two  inter 
secting  straight  lines  consists  of  two  perpendicular  lines  (I.  126,  25). 

What  is  the  locus  of  all  the  points  which  are  equally  distant  from  two 
parallel  lines? 


17.  Let  the  three  medial  lines  of  a  triangle  ABC 
meet  in  0.  Let  one  of  them,  AD,  be  produced  to  G, 
making  DG  =  DO,  and  join  CG.  Then,  the  sides  of 
the  triangle  OCG  are,  respectively,  two-thirds  of  the 
medial  lines  of  ABC  (I.  134). 

Also,  if  the  three  medial  lines  of  the  triangle  OCG 
be  drawn,  they  will  be  respectively  equal  to  J  AB,  \BG 


18.  In  any  triangle  ABC,  if  AD  is  drawn  perpen 
dicular  to  BC,  and  AE  bisecting  the  angle  BAG,  the 
angle  DAE  is  equal  to  one-half  the  difference  of  the 
angles  B  and  C  (I.  68). 


19.  If  BE  bisects  the  angle  B  of  a  triangle  ABC, 
and  CE bisects  the  exterior  angle  A  CD,  the  angle  E 
is  equal  to  one-half  the  angle  A.  B 


D  E 


20.  If  from  the  diagonal -.£7)  of  a  square  ABCD,  BE  is 
cut  off  equal  to  BC,  and  EF  is  drawn  perpendicular  to  BD, 
then,  DE=EF=FC. 


296 


EXEKCISES. 


21.  If  E  and  .Fare  the  middle  points  of  the  oppo 
site  sides,  AD,  BC,  of  a  parallelogram  ABCD,  the 
straight  lines  BE,  DF,  trisect  the  diagonal  AC. 


22.  The  sum  of  the  four  lines  drawn  to  the  vertices  of  a  quadrilateral  from 
any  point  except  the  intersection  of  the  diagonals,  is  greater  than  the  sum 
of  the  diagonals. 

D 

23.  The  straight  lines  joining  the  middle  points  of 
the  adjacent  sides  of  any  quadrilateral,  form  a  paral 
lelogram  whose  perimeter  is  equal  to  the  sum  of  the 
diagonals  of  the  quadrilateral  (I.  122). 


24.  The  intersection  of  the  straight  lines 
which  join  the  middle  points  of  opposite  sides 
of  any  quadrilateral,  is  the  middle  point  of  the 
straight  line  which  joins  the  middle  points  of 
the  diagonals  (I.  122,  108,  109). 


25.  The  four  bisectors  of  the  angles  of  a 
quadrilateral  form  a  second  quadrilateral,  the 
opposite  angles  of  which  are  supplementary. 


If  the  first  quadrilateral  is  a  parallelogram,  the  second  is  a  rectangle.     If 
the  first  is  a  rectangle,  the  second  is  a  square. 

A  ED 

26.  A  parallelogram  is  a  symmetrical  figure  with  re,.-* 
spect  to  its  centre  (intersection  of  the  diagonals),  (I.  140). 


27.  If  in  a  parallelogram  ABCD,  E  and  G  are 
any  two  symmetrical  points  in  the  sides  AD,  BC, 
and  F  and  H  any  two  symmetrical  points  in  the 
sides  AB,  DC,  the  figure  EFGff  is  a  parallelo 
gram  concentric  with  ABCD. 


28.  If  the  diameters  (I.  140)  EG,  FH,  joining  symmetrical    p 
points  in  the  opposite  sides  of  a  square  ABCD,  are  perpen 
dicular  to  each  other,  the  lines  joining  their  extremities  form 
a  second  square,  EFGJI,  concentric  with  ABCD. 


EXERCISES. 


297 


D 


PEOBLEM. 

29.  A  billiard-ball  is  placed  at  any  point  Pof  a  rectangular  table  ABCD. 
In  what  direction  must  it  be  struck  to  cause  it  to  return  to  the 
same  point  P,  after  impinging  successively  on  the  four  sides  of 
the  table,  the  ball,  before  and  after  impinging  on  a  side,  moving 
in  lines  which  make  equal  angles  with  the  side  ? 

What  is  the  length  of  the  whole  path  described  by  the  ball? 
Show  that  it  is  the  shortest  path  that  can  be  described  by  the 
ball  touching  the  four  sides  and  returning  to  the  same  point. 


GEOMETRY.— BOOK   II. 

THEOEEMS. 

30.  The  circle  is  a  symmetrical  figure  with  respect  to  any  diameter,  or 
with  respect  to  its  centre. 


31.  If  P  is  any  point  within  a  circle  whose  centre  is  0, 
and  APOB  is  the  diameter  through  P,  then,  AP  is  the 
least,  and  PB  the  greatest,  distance  from  P  to  the  cir 
cumference. 


^B/ 


32.  Prove  the  correctness  of  the  follow 
ing  method  of  drawing  a  tangent  to  a  given 
circumference  0,  from  a  given  point  A 
without  it. 

With  radius  A  0  and  centre  A,  describe 
an  arc  B  OB'.  With  centre  0,  and  radius 
equal  to  the  diameter  of  the  given  circle, 
describe  arcs  intersecting  the  first  in  B  and 
B'.  Join  OB,  OB',  intersecting  the 
given  circumference  in  T,  T'.  Then  AT, 
AT',  are  tangents. 


33.  The  bisectors  of  the  angles  contained  by 
the  opposite  sides  (produced)  of  an  inscribed 
quadrilateral  intersect  at  right  angles. 


298 


EXERCISES. 


34.  If  from  the  middle  point  A  of  an  arc  BC,  any 
chords  AD,  AE  are  drawn,  intersecting  the  chord  BC 
in  F  and  G,  FDEG  is  an  inscriptible  quadrilate 
ral.  (II.  99.) 


35.  If  A'B'C'  is  a  triangle  inscribed 
in  another  triangle  ABC,  the  three  cir 
cumferences  circumscribed  about  the  tri 
angles  AB'C',  BA'C',  CA'B',  inter 
sect  in  a  common  point  P. 

Let  P  be  the  intersection  of  two  of  the 
circumferences,  and  prove  that  the  third 
must  pass  through  P  (II.  99). 


36.  The  perpendiculars  from  the  angles  upon  the  opposite  sides  of  a  tri 
angle  are  the  bisectors  of  the  angles  of  the  triangle  formed  by  joining  the 
feet  of  the  perpendiculars  (II.  58,  99). 


37.  If  two  circumferences  are  tangent  internally,  and  the  radius  of  the 
larger  is  the  diameter  of  the  smaller,  then,  any  chord  of  the  larger  drawn 
from  the  point  of  contact  is  bisected  by  the  circumference  of  the  smaller. 
(II  15). 


38.  If  A  OB  is  any  given  angle  at  the  centre  of 
a  circle,  and  if  BC  can  be  drawn,  meeting  AO 
produced  in  C,  and  the  circumference  in  D,  so 
that  CD  shall  be  equal  to  the  radius  of  the  circle, 
then,  the  angle  C  will  be  equal  to  one-third  the 
angle  A  OB. 


Note.  There  is  no  method  known  of  drawing  BC,  under  these  conditions, 
and  with  the  use  of  straight  lines  and  circles  only,  A  OB  being  any  given 
angle :  so  that  the  trisection  of  an  angle,  in  general,  is  a  problem  that  cannot 
be  solved  by  elementary  geometry. 


39.  If  ABC  is  an  equilateral  triangle  inscribed  in 
a  circle,  and  P  any  point  in  the  arc  BC,  then  PA  = 
PB  +  PC. 


EXERCISES. 


299 


40.  If  a  triangle  ABC  is  formed  by  the 
intersection  of  three  tangents  to  a  circumfer 
ence,  two  of  which,  AM  and  AN,  are  fixed, 
while  the  third  BC  touches  the  circumference 
at  a  variable  point  P,  prove  that  the  perimeter 
of  the  triangle  ABCis  constant,  and  equal  to 
AM  +  AN,  or  2  AN. 

Also,  prove  that  the  angle  BOC  is  constant. 


41.  If  ABC  is  a  triangle  inscribed  in  a  circle,  and 
from  the  middle  point  D  of  the  arc  BC  a  perpen 
dicular  DE  is  drawn  to  AB ;  then,  (II.  57),  (1.87), 


If  the  perpendicular  D/E/  is  drawn  from  the 
middle  point  D'  of  the  arc  B A  C,  then  ^ 


Also  join  AD  and  draw  the  diameter  DD' ;  then, 
the  angle  ADD'  is  equal  to  one-half  the  difference  of  the  angles  ACB 'and 
ABC. 


42.  If  two  straight  lines  are  drawn  through  the  point  of  contact  of  two 
circles,  the  chords  of  the  intercepted  arcs  are  parallel. 


43.  Two  circles  are  tangent  internally  at  P,  and  a  chord  AB  of  the  larger 
circle  touches  the  smaller  at  C',  prove  that  PC  bisects  the  angle  APB 
(II.  62). 

E 


44.  If  through  P,  one  of  the  points  of  in 
tersection  of  two  circumferences,  any  two 
secants,  APB,  CP&  are  drawn,  the  straight  A 
lines,  AC,  DB,  joining  the  extremities  of  the 
secants,  make  a  constant  angle  E,  equal  to 
the  angle  MPN  formed  by  the  tangents  at  P. 


45.  If  through  one  of  the  points  of  intersection  of  two  circumferences,  a 
diameter  of  each  circle  is  drawn,  the  straight  line  which  joins  the  extremities 
of  these  diameters  passes  through  the  other  point  of  intersection,  is  parallel 
to  the  line  joining  the  centres,  and  is  ohogtcr-  than  any  other  line  drawn 
through  a  point  of  intersection  and  terminated  by  the  two  circumferences. 


300 


EXERCISES. 


46.  The  feet,  a,  &,  c,  of  the  perpendiculars  let  fall 
from  any  point  P  in  a  circumference  on  the  sides 
of  an  inscribed  triangle  ABC,  are  in  a  straight  line. 

Join  a&,  6c,  and  prove  that  the  angle  abC=  the 
angle  Abe  (II.  99). 


47.  If  equilateral  triangles  AB  C', 
EGA',  CAB',  are  constructed  on 
the  sides  of  any  triangle  ABC: 
1st.  The  circumferences  circumscrib 
ed  about  the  equilateral  triangles  in 
tersect  in  the  same  point  P;  2d.  The 
straight  lines  A  A',  BB',  CC',  are 
equal  and  intersect  in  P;  3d.  The 
centres  of  the  three  circumferences 
are  the  vertices  of  an  equilateral  tri 
angle  OO'O". 


48.  The  inscribed  and  the  three  escribed  circles  of  a  triangle  ABC 
being  drawn,  as  in  the  figure  on  p.  86,  let  D,  D',  D",  D'",  be  the  four 
points  of  contact  on  the  same  side  BC.  Designate  the  side  BC,  opposite 
to  the  angle  A,  by  a,  the  side  AC  by  6,  and  the  side  AB  by  c;  and  let 
s  —  \  (a  +  b  +  c).  Prove  the  following  properties : 

=  D'"D'  ==  6, 


=  CD"  =  s—a, 
BD  =  CD'  =  s—b, 
BD'  =  CD  =*-c,  D'"D"  =  b  +  c. 

Also,  let  a  circumference  be  circumscribed  about  the  triangle  ABC.  Prove 
that  this  circumference  bisects  each  of  the  six  lines  0  0',  0  0",  0  0"', 
0/  0",  0"  0'",  0'"  0' ;  and  that  the  points  of  bisection  are  the  centres 
of  circumferences  that  may  be  circumscribed  about  the  quadrilaterals 
BOG Wi  COAO",  AOBO'",  ABO'O",  BCO"0'",  CAO'"0',  re 
spectively. 

Finally,  designating  the  radius  of  the  circumscribed  circle  by  K;  the  radius 
of  the  inscribed  circle  by  r ;  the  radii  of  the  escribed  circles  by,  r',  r",  r"' ; 
the  perpendiculars  from  the  centre  of  the  circumscribed  circle  to  the  three 


sides  by  i/,  p",  p'" 


prove  the  following  relations: 


EXERCISES. 


301 


LOCI. 

49.  Find  the  locus  of  the  centre  of  a  circumference  which  passes  through 
two  given  points. 

50.  Find  the  locus  of  the  centre  of  a  circumference  which  is  tangent  to  two 
given  straight  lines. 

51.  Find  the  locus  of  the  centre  of  a  circumference  which  is  tangent  to 
a  given  straight  line  at  a  given  point  of  that  line,  or  to  a  given  circumfer 
ence  at  a  given  point  of  that  circumference. 

52.  Find  the  locus  of  the  centre  of  a  circumference  passing  through  a  given 
point  and  having  a  given  radius. 

53.  Find  the  locus  of  the  centre  of  a  circumference  tangent  to  a  given 
straight  line  and  having  a  given  radius. 

54.  Find  the  locus  of  the  centre  of  a  circumference  of  given  radius,  tan 
gent  externally  or  internally  to  a  given  circumference. 

c 


55.  A  straight  line  MN^  of  given  length,  is 
placed  with  its  extremities  on  two  given  perpen- 
dicular  lines,  AB,  CD]  find  the  locus  of  its 
middle  point  P  (Ex.  8). 


N 


56.  A  straight  line  of  given  length  is  inscribed  in  a  given  circle ;  find  the 
locus  of  its  middle  point. 


57.  A  straight  line  is  drawn  through  a  given  point 
J_,  intersecting  a  given  circumference  in  B  and  C] 
find  the  locus  of  the  middle  point,  P,  of  the  inter 
cepted  chord  B  C. 

Note  the  special  case  in  which  the  point  A  is  on 
the  given  circumference. 


58.  From  any  point  A  in  a  given  circumference,  a  straight  line  AP  of 
fixed  length  is  drawn  parallel  to  a  given  line  MN\  find  the  locus  of  the 
extremity  P. 

59.  Upon  a  given  base  BC,  a  triangle  ABO  is  constructed  having  a  given 
vertical  angle  A ;  find  the  locus  of  the  intersection  of  the  perpendiculars 
from  the  vertices  of  this  triangle  upon  the  opposite  sides  (II.  97). 

26 


302 


EXERCISES. 


60.  The  angle  A  CB  is  any  inscribed  angle  in  a 
given  segment  of  a  circle ;  A  C  is  produced  to  P, 
making  CP  equal  to  CB:  find  the  locus  of  P. 


61.  From  one  extremity  A  of  a  fixed  diameter 
AB,  any  chord  A  C  is  drawn,  and  at  C  a  tangent 
CD.  f  From  B,  a  perpendicular  BD  to  the  tan 
gent  is  drawn,  meeting  A  C  in  P.  Find  the  locus 
of  P. 


62.  A  triangle  ABC  is  given,  right  angled  at  A. 
Any  perpendicular,  EF,  to  BC,  is  drawn,  cutting  AB 
in  D,  and  CA  in  F.  Find  the  locus  of  Pt  the  inter 
section  of  BF  and  CD. 


63.  The  base  BC  of  a  triangle  is  given,  and  the  me 
dial  line  BE,  from  B,  is  of  a  given  length.  Find  the 
locus  of  the  vertex  A. 

Draw  A 0  parallel  to  EB.  Since  BO=BC,  0  is  a 
fixed  point ;  and  since  A0  =  2 BE,  OA  is  a  constant,, 
distance. 


64.  An  angle  BAG  is  given  in  position,  and 
points   B  and   C  are  taken  in  its  sides  so  that 
AB  +  AC  shall  be  a  given  constant  length.    Find 
the  locus  of  the  centre  of  the  circle  circumscribed 
about  the  triangle  ABC  (Ex.  41). 

Also,  if  the  points  B  and  C  are  so  taken  that 
AB  —  A  C  is  a  given  constant  length,  find  the  locus 
of  the  centre  of  the  circle  circumscribing  ABC 
(Ex.  41). 

Also  find  the  locus  of  the  middle  point  of  BC. 

65.  The  base  BC  of  a  triangle  ABC  is  given  in  position,  and  the  vertical 


EXERCISES.  303 

angle  A  is  of  a  given  magnitude  ;  find  the  loci  of  the  centres  of  the  inscribed 
and  escribed  circles- 
In  the  figure,  p.  86,  we  have  the  angles  JBOC=W°  +  $A,  BO'C  = 


The  loci  are  circumferences  whose  centres  lie  in  the  circumference  of  the 
circle  circumscribed  about  ABC. 

66.  Find  the  locus  of  all  the  points  the  sum  of  the  distances  of  each  of 
which  from  two  given  straight  lines  is  a  given  constant  length  (Ex.  13). 

Show  that  the  locus  consists  of  four  straight  lines  forming  a  rectangle. 

67.  Find  the  locus  of  all  the  points  the  difference  of  the  distances  of  each 
of  which  from  two  given  straight  lines  is  a  given  constant  length  (Ex.  13). 

Show  that  the  locus  consists  of  parts  of  four  straight  lines  whose  intersec 
tions  form  a  rectangle. 

68.  If  in  Ex.  66  by  sum  is  understood  algebraic  sum,  and  distances  falling 
on  opposite  sides  of  the  same  line  have  opposite  algebraic  signs,  show  that 
Ex.  66  includes  Ex.  67,  and  the  locus  consists  of  the  whole  of  four  indefinite 
lines  whose  intersections  form  a  rectangle. 

PEOBLEMS. 

The  most  useful  general  precept  that  can  be  given,  to  aid  the  student  in 
his  search  for  the  solution  of  a  problem,  is  the  following.  Suppose  the 
problem  solved,  and  construct  a  figure  accordingly  :  study  the  properties  of 
this  figure,  drawing  auxiliary  lines  when  necessary,  and  endeavor  to  discover 
the  dependence  of  the  problem  upon  previously  solved  problems.  This  is  an 
analysis  of  the  problem.  The  reverse  process,  or  synthesis,  then  furnishes  a 
construction  of  the  problem.  In  the  analysis,  the  student's  ingenuity  will  be 
exercised  especially  in  drawing  useful  auxiliary  lines  ;  in  the  synthesis,  he  will 
often  find  room  for  invention  in  combining  in  the  most  simple  form  the 
several  steps  suggested  by  the  analysis. 

The  analysis  frequently  leads  to  the  solution  of  a  problem  by  the  intersec 
tion  of  loci.  The  solution  may  turn  upon  the  determination  of  the  position 
of  a  particular  point.  By  one  condition  of  the  problem  it  may  appear  that 
this  required  point  is  necessarily  one  of  the  points  of  a  certain  line  ;  this  line 
is  a  locus  of  the  point  satisfying  that  condition.  A  second  condition  of  the 
problem  may  furnish  a  second  locus  of  the  point  ;  and  the  point  is  then  fully 
determined,  being  the  intersection  of  the  two  loci. 

Some  of  the  following  problems  are  accompanied  by  an  analysis  to  illus 
trate  the  process. 

69.  A  triangle  ABC  being  given,  to  draw  DE  parallel 
to  the  base  BG  so  that  DE=  DB  +  EC. 

Analysis.  Suppose  the  problem  solved,  and  that  DE  is 
the  required  parallel.  Since  DE=DB  +  EC,  we  may 
divide  it  into  two  portions,  DP  and  PE,  respectively  equal 
to  DB  and  EG.  Join  PB,  PC.  Then  we  have  the 
angle  DBP=  DPB  =  PBC;  therefore,  the  line  PB  bi 
sects  the  angle  ABC.  In  the  same  manner  it  is  shown  that  CP  bisects 


304 


EXERCISES. 


the  angle  ACB.  The  point  P,  then,  lies  in  each  of  the  bisectors  of  the 
base  angles  of  the  triangle,  and  is  therefore  the  intersection  of  these  bisectors. 
Hence  we  derive  the  following  construction. 

Construction.  Bisect  the  angles  B  and  C  by  straight  lines.  Through  the 
intersection  P  of  the  bisectors,  draw  the  line  DPE  parallel  to  BC.  This 
line  satisfies  the  conditions.  For  we  have,  by  the  construction,  the  angle 
DBP=  PBC=  BPD;  therefore,  PD  =  DB]  and  in  the  same  manner, 
PE=EC\  hence,  DE  =  DB  +  EG. 


We  have,  however,  tacitly  assumed  that  DE  is 
to  be  drawn  so  as  to  cut  the  sides  of  the  triangle 
ABC  between  the  vertex  and  the  base.  Suppose 
it  drawn  cutting  AB  and  AC  produced.  Then  the 
same  analysis  shows  that  the  point  P  is  found  by 
bisecting  the  exterior  angles  CBD,  BCE.  Thus 
the  problem  has  two  solutions,  if  the  position  of 
DE  is  not  limited  to  one  side  of  the  base  BC. 


70.  To  determine  a  point  whose 
distances  from  two  given  inter 
secting  straight  lines,  AB,  A/B/, 
are  given. 

Analysis.  The  locus  of  all  the 
points  which  are  at  a  given  dis 
tance  from  AB  consists  of  two 
parallels  to  AB,  CE  and  DP, 

each  at  the  given  distance  from  AB.  The  locus  of  all  the  points  at  a  given 
distance  from  A/B/  consists  of  two  parallels,  C/E/  and  D'F',  each  at  the 
given  distance  from  A'B'.  The  required  point  must  be  in  both  loci,  and 
therefore  in  their  intersection.  There  are  in  this  case  four  intersections  of 
the  loci,  and  the  problem  has  four  solutions. 

Construction.  At  any  point  of  AB,  as  A,  erect  a  perpendicular  CD,  and 
make  AC  =  AD  =  the  given  distance  from  AB ;  through  C  and  D  draw 
parallels  to  AB.  In  the  same  manner,  draw  parallels  to  A/B/  at  the  given 
distance  A/C/  —  A' If.  The  intersection  of  the  four  parallels  determines 
the  four  points  Pi,  P2,  P3,  P4,  each  of  which  satisfies  the  conditions. 

E  c 


M 


71.  Given  two  perpendiculars,  AB  and  CD,  inter 
secting  in  0.  To  construct  a  square,  one  of  whose 
angles  shall  coincide  with  one  of  the  right  angles  at  0, 
and  the  vertex  of  the  opposite  angle  of  the  square 
shall  lie  on  a  given  straight  line  EF.  (Two  solutions. ) 


EXEECISES. 


72.  In  a  given  rhombus,  ABCD,  to  inscribe  B 

a  square  ^mK  (Ex.71.) 


305 


73.  In  a  given  straight  line,  to  find  a  point  equally  distant  from  two  given 
points  without  the  line. 

74.  To  construct  a  square,  given  its  diagonal. 

75.  In  a  given  square,  to  inscribe  a  square  of  given  magnitude. 

76.  From  two  given  points  on  the  same  side  of  a  given  straight  line,  to 
draw  two  straight  lines  meeting  in  a  given  straight  line  and  making  equal 
angles  with  it.  (Ex.  4. ) 

77.  Through  a  given  point  P  within  a  given  angle,  to  draw  a  straight  line, 
terminated  by  the  sides  of  the  angle,  which  shall  be  bisected  at  P. 

78.  Given  .two  straight  lines  which  cannot  be  produced  to  their  intersec 
tion,  to  draw  a  third  which  would  pass  through  their  intersection  and  bisect 
their  contained  angle. 

79.  Through  a  given  point,  to  draw  a  straight  line  making  equal  angles 
with  two  given  straight  lines. 

80.  Given  the  middle  point  of  a  chord  in  a  given  circle,  to  draw  the 
chord. 

81.  To  draw  a  tangent  to  a  given  circle  which  shall  be  parallel  to  a  given 
straight  line. 

82.  In  the  prolongation  of  any  given  chord  AB  of  a  circle,  to  find  a  point 
P,  such  that  the  tangent  PT,  drawn  from  it  to  the  circle,  shall  be  of  a  given 
length. 

83.  To  draw  a  tangent  to  a  given  circle,  such  that  its  segment  intercepted 
between  the  point  of  contact  and  a  given  straight  line  shall  have  a  given 
length. 

In  general,  there  are  four  solutions.     Show  when  there  will  be  but  three 
solutions,  and  when  but  two ;  also,  when  no  solution  is  possible. 

84.  Through  a  given  point  within  or  without  a  given  circle,  to  draw  a 
straight  line,  intersecting  the  circumference,  so  that  the  intercepted  chord 
shall  have  a  given  length.     (Two  solutions. ) 

85.  Through  a  given  point,  to  draw  a  straight  line,  intersecting  two  given 
circumferences,  so  that  the  portion  of  it  intercepted  between  the  circumfer 
ences  shall  have  a  given  length.    (Two  solutions. ) 

86.  In  a  given  circle,  inscribe  a  chord  of  a  given  length  which  produced 
shall  be  tangent  to  another  given  circle. 

87.  Construct  an  angle  of  60°,  one  of  120°,  one  of  30°,  one  of  150°,  one 
of  45°  and  one  of  135°. 

88.  To  find  a  point  within  a  given  triangle,  such  that  the  three  straight 
lines  drawn  from  it  to  the  vertices  of  the  triangle  shall  make  three  equal 
angles  with  each  other. 

When  will  the  problem  be  impossible  ? 

89.  Construct  a  parallelogram,  given,  1st,  two  adjacent  sides  and  one  diago- 

26*  U 


306 


EXERCISES. 


nal ;  2d,  one  side  and  the  diagonals ;  3d,  the  diagonals  and  the  angle  they 
contain. 

90.  Construct  a  triangle,  given  the  base,  the  angle  opposite  to  the  base, 
and  the  altitude. 

Analysis.  Suppose  BAG  to  be  the  required  tri 
angle.  The  side  BC  being  fixed  in  position  and 
magnitude,  the  vertex  A  is  to  be  determined.  One 
locus  of  A  is  an  arc  of  a  segment,  described  upon 
AB,  containing  the  given  angle.  Another  locus  of 
A  is  a  straight  line  MN  drawn  parallel  to  BC,  at  a  distance  from  it  equal 
to  the  given  altitude.  Hence  the  position  of  A  will  be  found  by  the  inter 
section  of  these  two  loci,  both  of  which  are  readily  constructed. 

Limitation.  If  the  given  altitude  were  greater  than  the  perpendicular 
distance  from  the  middle  of  BC  to  the  arc  BAG,  the  arc  would  not  inter 
sect  the  line  MN,  and  there  would  be  no  solution  possible. 

The  limits  of  the  data,  within  which  the  solution  of  any  problem  is  pos 
sible,  should  always  be  determined. 

91.  Construct  a  triangle,  given  the  base,  the  medial  line  to  the  base,  and 
the  angle  opposite  to  the  base. 

92.  Construct  a  triangle,  given  the  base,  an  angle  at  the  base,  and  the  sum 
or  difference  of  the  other  two  sides. 

Analysis.  On  the  sides  of  "the  given  angle,  B,  take 
BG  ~  given  base,  and  BD  =  given  sum  or  difference 
of  the  sides.  Join  CD.  The  problem  is  reduced  to 
drawing,  from  (7,  a  line  CA,  which  shall  cut  BD,  or 
BD  produced,  in  a  point  A,  so  that  GA  shall  be  equal 
to  AD,  which  is  obviously  effected  by  making  the  angle  DCA  =  the  angle 
ADC. 

If,  when  the  difference  of  AB  and  AC  is  given,  AC  is  to  be  the  greater 
side,  BD  =  AC  —  AB  is  to  be  taken  in  AB  produced  through  B. 

93.  Construct  a  triangle,  given  the  base,  the  angle  opposite  to  the  base, 
and  the  sum  or  difference  of  the  other  two  sides. 

Analysis.    Suppose  ABC  is  the  required  triangle. 
First,  when  the  sum  of  AB  and  AC  is  given,  produce 
BA  to  D,  making  BD  =  AB  +  A  C.   Join  CD.    The 
angle  D  is  one-half  of  B  A  C,  and  is  therefore  known. 
Hence  the  following  construction.      Make  an  angle 
BDC   equal    to  one-half  the    given    angle.       Take 
DB  —  given  sum  of  sides.     From  B  as  centre,  and  with  radius  equal  to  the 
given  base,  describe  an  arc  cutting  DC  in  C.     Draw   CA,  making  the 
angle  DCA  =  the  angle  BDC. 

Secondly,  when  the  difference  of  AB  and  AC  is  given;  take  BD/ '  — 
AB  —  AC,  and  join  CD'.  The  angle  AD'C  is  one-half  the  supplement 
of  BA  C,  and  hence  the  construction  can  readily  be  found. 

This  problem  can  also  be  solved  by  an  application  of  Ex.  41. 


EXERCISES.  307 

94.  Construct  a  triangle,  given  the  base,  the  sum  or  difference  of  the  other 
two  sides,  and  the  difference  of  the  angles  at  the  base. 

Analysis.  In  the  preceding  figure,  the  angle  BCD'  =  J  (ACB  —  ABC], 
and  BCD  =  90°  +  BCD' ;  and  hence  a  construction  can  readily  be  found. 

95.  Construct  a  triangle,  given,  1st,  two  angles  and  the  sum  of  two  sides ; 
2d,  two  angles  and  the  perimeter. 

96.  Construct  a  triangle,  given, 

1st.  Two  sides  and  one  medial  line 
2d.  One  side  and  two  medial  lines ; 
3d.  The  three  medial  lines. 
See  Exercise  17. 

97.  Construct  a  triangle,  given  an  angle,  the  bisector  of  that  angle,  and 
the  perpendicular  from  that  angle  to  the  opposite  side. 

98.  Construct  a  triangle,  given  the  middle  points  of  its  sides. 

99.  Construct  a  triangle,  given  the  feet  of  the  perpendiculars  from  the 
angles  on  the  opposite  sides.  (Ex.  36. ) 

100.  Construct  a  triangle,    given  the  perimeter,    one    angle,   and    one 
altitude. 

101.  Construct  a  triangle,  given  an  angle,  together  with  the  medial  line' 
and  the  perpendicular  from  that  angle  to  the  opposite  side. 

102.  Construct  a  triangle,  given  the  base,  the  sum  or  the  difference  of  the 
other  two  sides,  and  the  radius  of  the  inscribed  circle.  (Ex.  48. ) 

103.  Construct  a  triangle,  given  the  centres  of  the  three  escribed  circles. 

104.  Construct  a  triangle  having  its  vertices  on  two  given  concentric  cir 
cumferences,  its  angles  being  given. 

105.  Divide  a  given  arc  into  two  parts  such  that  the  sum  of  their  chords 
shall  be  a  given  length.  (Ex.  41.) 

106.  Construct  a  square,  given  the  sum  or  the  difference  of  its  diagonal 
and  side.  (Ex.  20. ) 

107.  With  a  given  radius,  describe  a  circumference,  1st,  tangent  to  two 
given  straight  lines ;  2d,  tangent  to  a  given  straight  line  an'd  to  a  given  cir 
cumference  ;  3d,  tangent  to  two  given  circumferences ;  4th,  passing  through 
a  given  point  and  tangent  to  a  given  straight  line ;  5th,  passing  through  a 
given  point  and  tangent  to  a  given  circumference ;  6th,  having  its  centre  on  a 
given  straight  line,  or  a  given  circumference,  and  tangent  to  a  given  straight 
line,  or  to  a  given  circumference.   (Exs.  52,  53,  54. ) 

108.  Describe  a  circumference,  1st,  tangent  to  two  given  straight  lines,  and 
touching  one  of  them  at  a  given  point  (Exs.  50,  51) ;  2d,  tangent  to  a  given 
circumference  at  a  given  point  and  tangent  to  a  given  straight  line  ;  3d,  tan 
gent  to  a  given  straight  line  at  a  given  point  and  tangent  to  a  given  cir 
cumference  (Ex.  51) ;  4th,  passing  through  a  given  point  and  tangent  to  a 
given  straight  line  at  a  given  point ;  5th,  passing  through  a  given  point  and 
tangent  to  a  given  circumference  at  a  given  point. 

109.  Draw  a  straight  line  equally  distant  from  three  given  points. 
When  will  there  be  but  three  solutions,  and  when  an  indefinite  number  of 

solutions  ? 


308  EXEKCISES. 

110.  Describe  a  circumference  equally  distant  from  four  given  points;  the 
distance  from  a  point  to  the  circumference  being  measured  on  a  radius,  or 
radius  produced. 

In  general  there  are  four  solutions.  If  three  of  the  given  points  are  in  a 
straight  line,  one  of  the  four  circumferences  becomes  a  straight  line. 


111.  An  angle  A  is  given  in  position,  and  a  point 
P  in  its  plane.  It  is  required  to  draw  a  straight  line 
through  P,  intersecting  the  sides  of  the  angle  and 
forming  a  triangle  ABC  of  a  given  perimeter. 
(Ex.  40.) 


112.  With  a  given  point  as  a  centre,  describe  a  circle  which  shall  be 
divided  by  a  given  straight  line  into  segments  containing  given  angles. 

113.  Through  a  given  point  without  a  given  circle,  draw  a  secant,  so  that 
the  portion  of  it  without  the  circle  shall  be  equal  to  the  portion  within. 
(Ex.  96.) 


E 


114.  Inscribe  a  straight  line  MN,  of  given 
length,  between  two  given  straight  lines  A  B, 
CD,  and  parallel  to  a  given  straight  line  EF. 


115.  Inscribe  a  straight  line  of  given  length  between  two  given  circumfer 
ences,  and  parallel  to  a  given  straight  line. 

116.  Through  P,  one  of  the  points  of  intersection  of  two  circumferences, 
draw  a  straight  line,  terminated  by  the  circumferences,  which  shall  be  bi 
sected  in  P. 

117.  Through  one  of  the  points  of  intersection  of  two  circumferences, 
draw  a  straight  line,  terminated  by  the  circumferences,  which  shall  have  a 
given  length. 

118.  Given  two  parallels  AB,  CD,  and  two  other  parallels  A'B',  C'D', 
inclined  to  the  first ;  through  a  given  point  P,  in  their  plane,  draw  a  straight 
line  such  that  the  portion  of  it  intercepted  between  the  parallels  AB,  CD, 
shall  be  equal  to  the  portion  of  it  intercepted  between  the  parallels  A'B', 
C'D'.  (Ex.  77.) 

119.  From  two  given  points,  A,  B,  on  the  same  side  of  a  given  straight 
line  MN,  draw,  straight  lines,  meeting  in  a  point  P  of  MN,  so  that  the  angle 
APM  shall  be  equal  to  double  the  angle  BPN. 

Analysis.  The  solution  of  Exercise  76,  suggests  the  possible  advantage  of 


EXERCISES. 


309 


employing  the  symmetrical  point  of  one  of  the  given  points.  Let  B'  be 
the  symmetrical  point  of  B  with  respect 
to  MN  (I.  1 35).  Join  B'P  and  produce 
it  toward  E.  Then,  since  APM=*  M~ 
2  BPN=  2  B'PN  =  2  MPE,  B'PE  bi 
sects  the  angle  APM.  Therefore,  the 
problem  is  reduced  to  finding  a  point 
Pm  MN  such  that  B'PE  shall  bisect  the  angle  APM.  With  B'  as  centre 
describe  an  arc  through  A,  cutting  MN  in  C.  The  perpendicular  B'E  upon 
AC,  evidently  intersects  MN  in  the  required  point. 

120.  With  the  vertices  of  a  given  triangle  as  centres,  describe  three  cir 
cumferences  each  of  which  shall  be  tangent  to  the  other  two.    (Four  solu 
tions.)     (Ex.48.) 

121.  Construct  a  quadrilateral,  given  its  four  sides  and  the  straight  line 
joining  the  middle  points  of  two  opposite  sides.   (Ex.  24. ) 

122.  Construct  a  pentagon,  given  the  middle  points  of  its  sides. 

The  middle  points  of  all  the  diagonals  can  be  determined  by  the  principle 
of  Ex.  23. 

123.  Find  a  point  in  a  given  straight  line,  such  that  tangents  drawn  from 
it  to  two  given  circumferences  shall  make  equal  angles  with  the  line.     (Four 
solutions:)     Compare  Ex.  76. 

124.  If  a  figure  is  moved  in  a  plane,  it  may  be  brought  from  one  position 
to  any  other,  by  revolving  it  about  a  certain  fixed  point  (that  is,  by  causing 
each  point  of  the  figure  to  move  in  the  circumference  of  a  circle  whose  centre 
is  the  fixed  point).     Find  that  point,  for  two  given  positions  of  the  figure. 


GEOMETRY.— BOOK  III.       . 

THEOREMS. 

125.  If  three  parallels  AA',  BB',  CC',  intercept  on  two 
straight  lines  AC,  A'C',  the  segments  AB  and  BG,  or 
A/B/  and  B/  C',  in  a  given  ratio  m  :  n,  that  is,  if 

AB:  BC^A'B'  :  B'C'  =  m:n; 

then,  (m  +  n).  BB'  =  n.AA'  +  m.  CC'. 

(III.  25,  10.) 

126.  In  any  triangle  ABC,  if  from  the  vertex  A, 
AE  is  drawn  to  the  circumference  of  the  circumscribed 
circle,  and  AD  to  the  base  BC,  making  the  angles 
CAE  and  BAD  equal  to  each  other,  then  (III.  25), 

ABXAC=ADXAE. 


310  EXERCISES. 

127.  From  the  preceding  theorem,  deduce  as  a  corol 
lary  the  following:  In  any  triangle  ABC,  if  from  the 
vertex  A,  AE  is  drawn  bisecting  the  angle  A,  meeting 
the  circumference  of  the  circumscribed  circle  in  E  and 
the  base  EC  in  D,  then 

ABXAC=ADXAE. 
Also  deduce  (III.  65). 

128.  If  ABCD  is  a  given  parallelogram,  and  AN  a  variable  straight  line 
drawn  through  A  cutting  BC  in  M  and  CD  in  N  ';   then,   the  product 
EM.  DN  is  constant.    (III.  25.  ) 

129.  If  ABCD  is  any  parallelogram,  and  from  any  point  P  in  the  diago 
nal  A  C  (or  in  AC  produced)  PM  is  drawn  cutting  BA  in  M,  BC  in  N, 
AD  in  M'  and  DC  in  TV'  ;  then,  PM.PN=  PM'.PN'.  (III.  25.) 

130.  If  a  square  DEFG  is  inscribed  in  a  right  triangle  ABC,  so  that  a 
side  DE  coincides  with  the  hypotenuse  EG  (the  vertices  F  and  G  being  in 
the  sides  AC  and  JLB)  ;  then,  the  side  DEis  a  mean  proportional  between 
the  segments  BD  and  EC  of  the  hypotenuse.   (III.  25.  ) 

131.  If  a  straight  line  AE  is  divided  at  C  and  D  so  that  AB.AD  = 
J.6T2,  and  if  from  J.  any  straight  line  AEls  drawn  equal  to  AC]  then,  EC 
bisects  the  angle  DEB.  (III.  10,  32,  23.) 

132.  If  a,  6,  c,  denote  the  three  perpendiculars  from  the  three  vertices  of 
a  triangle  upon  any  straight  line  MN  in  its  plane,  and  p  the  perpendicular 
from  the  intersection  of  the  three  medial  lines  of  the  triangle  upon  the  same 
straight  line  MN]  then,  (Ex.  125,) 


133.  If  ABC  and  A'BC  are  two  triangles  having  a  common  base  '  E  C 
and  their  vertices  in  a  line  A  A'  parallel  to  the  base,  and  if  any  parallel  to 
the  base  cuts  the  sides  AB  and  AC  in  D  and  E,  and  the  sides  A  'B  and 
A'C  in  jy  and  E/  ;  then  DE  =  D'E'. 

134.  If  two  sides  of  a  triangle  are  divided  proportionally,  the  straight 
lines  drawn  from  corresponding  points  of  section  to  the  opposite  angles  inter 
sect  on  the  line  joining  the  vertex  of  the  third  angle  and  the  middle  of  the 
third  side. 

135.  The  difference  of  the  squares  of  two  sides  of  any  triangle  is  equal  to 
the  difference  of  the  squares  of  the  projections  of  these  sides  on  the  third 
side.  (III.  48.) 

136.  If  from  any  point  in  the  plane  of  a  polygon,  perpendiculars  are  drawn 
to  all  the  sides,  the  two  sums  of  the  squares  of  the  alternate  segments  of  the 
sides  are  equal.   (Ex.135.) 

137.  If  0  is  the  centre  of  the  circle  circumscribed  about  a  triangle  ABC, 
and  P  is  the  intersection  of  the  perpendiculars  from  the  angles  upon  the 
opposite  sides  ;  the  perpendicular  from  0  upon  the  side  BC  is  equal  to  one- 
half  the  distance  AP.  (I.  132),  (III.  25,  30.) 


E.XEKCISES.  311 

138.  In  any  triangle,  the  centre  of  the  circumscribed  circle,  the  intersec 
tion  of  the  medial  lines,  and  the  intersection  of  the  perpendiculars  from  the 
angles  upon  the  opposite  sides,  are  in  the  same  straight  line  ;  and  the  dis 
tance  of  the  first  point  from  the  second  is  one-half  the  distance  of  the  second 
from  the  third. 

139.  If  d  denotes  the  distance  of  a  point  P  from  the  centre  of  a  circle, 
and  r  the  radius  ;  and  if  any  straight  line  drawn  through  P  cuts  the  cir 
cumference  in  the  points  A  and  B]  then,  the  product  PA.PB  is  equal  to 
r2  —  d3  if  Pis  within  the  circle,  and  to  d2  —  r2  if  Pis  without  the  circle. 

140.  In  any  quadrilateral,  the  sum  of  the  squares  of  the  diagonals  is  equal 
to  twice  the  sum  of  the  squares  of  the  straight  lines  joining  the  middle  points 
of  the  opposite  sides.  (III.  64)  and  (Ex.  23.  ) 

141.  In  any  quadrilateral  ABCD  inscribed  in  a  cir- 
cle,  the  product  of  the  diagonals  is  equal  to  the  sum 
of  the  products  of  the  opposite  sides  ;  that  is, 


(Make  the  angle  DAE  '  =  BAG,  and  prove  that 
AD.BC=AC.DE,  and  AB.DC  =  AC.BE.} 

142.  In  an  inscribed  quadrilateral  ABCD,  if  F  is  the  intersection  of  the 
diagonals  A  C  and  BD  ;  then 

AD.AB_AF  ,m        , 

CB.CD~  FG'  Ull.  65.  J 

143.  In  an  inscribed  quadrilateral  ABCD, 

AD.AB+CB.CD  =  AC 
BA.BC+DA.DC      BD' 

144.  In  an  inscribed  quadrilateral,  the  product  of  the  perpendiculars  let 
fall  from  any  point  of  the  circumference  upon  two  opposite  sides  is  equal  to 
the  product  of  the  perpendiculars  let  fall  from  the  same  point  upon  the  other 
two  sides.   (III.  65.) 

145.  If  from  a  point  P  in  a  circumference, 
any  chords  PA,  PB,  PC,  are  drawn,  and  any 
straight  line  MN  parallel  to  the  tangent  at  P, 
cutting  the  chords  (or  chords  produced)  in  a, 
b,  c;    then,  the  products  PA.  Pa,   PB.Pb, 
PC.  PC,  are  equal. 

146.  If  two  tangents  are  drawn  to  a  circle  at  the  extremities  of  a  diameter, 
the  portion  of  any  third  tangent  intercepted  between  them  is  divided  at  its 
point  of  contact  into  segments  whose  product  is  equal  to  the  square  of  the 
radius. 

147.  If  on  a  diameter  of  a  circle  two  points  are  taken  equally  distant  from 
the  centre,  the  sum  of  the  squares  of  the  distances  of  any  point  of  the  cir 
cumference  from  these  two  points  is  constant. 


312  EXERCISES. 

148.  If  a  point  P  on  the  circumference  of  a  circle  is  taken  as  the  centre 
of  a  second  circle,  and  any  tangent  is  drawn  to  the  second  circle  cutting  the 
first  in  M  and  N;  then,  the  product  PM.PN  is  constant. 

149.  The  perpendicular  from  any  point  of  a  circumference  upon  a  chord  is 
a  mean  proportional  between  the  perpendiculars  from  the  same  point  upon 
the  tangents  drawn  at  the  extremities  of  the  chord. 

150.  If  AB  is  the  chord  of  a  quadrant  of  a  circle  whose  centre  is  0,  and 
any  chord  MN  parallel  to  AB  cuts  the  radii  OA,  OB  in  P  and  Q ;  then 

MP2  +  PN2  =  AB2.         (III.  48)  and  (Ex.  139.) 

151.  If  ABCD  is  any  parallelogram,  and  any  circumference  is  described 
passing  through  A  and  cutting  AB,  AC,  AD,  in  the  points  F,  G,  H,  re 
spectively;  then 

AF.AB  +  AH.AD  =  AG.AC. 

152.  In  any  isosceles  triangle,  the  square  of  one  of  the  equal  sides  is  equal 
to  the  square  of  any  straight  line  drawn  from  the  vertex  to  the  base  plus  the 
product  of  the  segments  of  the  base. 

153.  If  A  and  B  are  the  centres  of  two  circles  which  touch  at  C,  and  Pis 
a  point  at  which  the  angles  APC  and  BPC  are  equal,  and  if  from  P  tan 
gents  PD  and  PE  are  drawn  to  the  two  circles ;  then, 

PD.PE  =  PC2.  (HI.  21  and  66.) 

154.  If  two  circles  touch  each  other,  secants  drawn  through  their  point  of 
contact  and  terminating  in  the  two  circumferences  are  divided  proportionally 
at  the  point  of  contact. 

155.  If  two  circles  are  tangent  externally,  the  portions  of  their  common 
tangent  included  between  the  points  of  contact  is  a  mean  proportional  be 
tween  the  diameters  of  the  circles. 

156.  Two  circles  are  tangent  internally  at  A,  and  from  any  point  Pin  the 
circumference  of  the  exterior  circle  a  tangent  PM  is  drawn  to  the  interior 
circle  ;  prove  that  the  ratio  PA  :  PM  is  constant. 

157.  If  two  circles  intersect  in  the  points  A  and  B,  and  through  A  any 
secant  CAD  is  drawn   terminated   by  the   circumferences  at  C  and  D, 
the  straight  lines  BC  and  BD  are  to  each  other  as  the  diameters  of  the 
circles. 

158.  If  a  fixed  circumference  is  cut  by  any  circumference  which  passes 
through  two  fixed  points,  the  common  chord  passes  through  a  fixed  point. 

159.  Two  chords  AB  and  CD,  perpendicular  to  each  other,  intersect  in  a 
point  P  either  within  or  without  the  circle,  and  the  line  OP  is  drawn  from 
the  centre  0.     Prove  that  if  D  is  the  diameter  of  the  circle, 

PA2  +  PB2  +  PC*  +  PD*  =  D*, 
and 


EXERCISES.  313 

160.  If  any  number  of  circumferences  pass  through  the  same  two  points, 
and  if  through  one  of  these  points  any  two  straight  lines  are  drawn,  the  cor 
responding  segments  of  these  lines  intercepted  between  the  circumferences 
are  proportional. 

161.  If  a  triangle  ABC  is  inscribed  in  a  circle,  and  from  the  vertex  A, 
AD  and  AE  are  drawn  parallel  to  the  tangents  at  B  and  C  respectively  and 
cutting  the  base  BC  in  D  and  E;  then 


£D:J)E=AB2  :  AC*. 

162.  Let  AB  be  a  given  straight  line.    At  A  erect 
the  perpendicular  AD  =  AB  ;  in  BA  produced  take 
A0  =  ?  AB  ;  with  centre  0  and  radius  OD  describe 

a  circumference,  cutting  AB  and  AB  produced  in  C      c  o    A      c  B 

and  C/  ;  prove  that  AB  is  divided  in  extreme  and 
mean  ratio,  internally  at  C,  and  externally  at  C'. 

163.  If  a  rhombus  ABCD  is  circumscribed  about  a  circle,  any  tangent 
MN  determines  on  two  adjacent  sides  AB,  AD,  two  segments  BM,  DN, 
whose  product  is  constant. 

164.  If  in  a  semicircle  whose  diameter  is  AB,  any  two  chords  AC  and 
B  D  are  drawn  intersecting  in  P,  then 

AB*  =  AC.AP  +  BD.BP. 

165.  If  0  is  the  intersection  of  the  three  medial  lines  of  a  triangle  ABC, 
prove  the  relations 

=  3  (OA2  +  ~OJ32  +  Otf  2), 
3  OB*  =  AB 


166.  If  0  is  the  intersection  of  the  three  medial  lines  of  a  triangle  ABC, 
and  P  any  point  in  the  plane  of  the  triangle  ;  then, 


167.  If  R,  r,  r',  r//,  r/x/,  are  respectively  the  radii  of  the  circumscribed, 
the  inscribed,  and  the  three  escribed  circles  in  any  triangle,  and  if  d,  d',  d", 
d"',  are  respectively  the  distances  from  the  centre  of  the  circumscribed 
circle  to  the  centres  of  the  inscribed  and  escribed  circles,  prove  the  relations 

=  d/  2  —  2Rr'  =  d"  2  —  2  Rr"  =  d'"  —  2 


12 

27 


314 


EXERCISES. 


LOCI. 


168.  From  a  fixed  point  0,  a  straight  line  OA  is  drawn 
to  any  point  in  a  given  straight  line  MN,  and  divided  at 
P  in  a  given  ratio  m  :  n  (i.  e.,  so  that  OP :  PA  =  m:n); 
find  the  locus  of  P. 


169.  From  a  fixed  point  0,  a  straight  line  OA 
is  drawn  to  any  point  in  a  given  circumference, 
and  divided  at  P  in  a  given  ratio ;  find  the  locus 
of  P. 

170.  Find  the  locus  of  a  point  whose  distances  from  two  given  straight 
lines  are  in  a  given  ratio.     (The  locus  consists  of  two  straight  lines. ) 

171.  Find  the  locus  of  the  points  which  divide  the  various  chords  of  a 
given  circle  into  segments  (external  or  internal)  whose  product  is  equal  to  a 
given  constant,  k2.  (III.  56,  59.) 

172.  Find  the  locus  of  a  point  the  sum  of  whose  distances  from  two  given 
straight  lines  is  equal  to  a  given  constant  k. 

173.  Find  the  locus  of  a  point  the  difference  of  whose  distances  from  two 
given  straight  lines  is  equal  to  a  given  constant  k. 

174.  Find  the  locus  of  a  point  such  that  the  sum  of  the  squares  of  its  dis 
tances  from  two  given  points  is  equal  to  a  given  constant,  k2.  (III.  62.) 

175.  Find  the  locus  of  a  point  such  that  the  difference  of  the  squares  of 
its  distances  from  two  given  points  is  equal  to  a  given  constant  k2.  (III.  62. ) 

176.  If  A,  B  and  C  are  three  given  points  in  the  same  straight  line,  find 
the  locus  of  a  point  P  at  which  the  angles  APB  and  BPG  are  equal. 
(Ex.  131.) 

177.  Through  J.,  one  of  the  points  of  intersection  of  two  given  circles, 
any  secant  is  drawn  cutting  the  two  circumferences  in  the  points  B  and  C ; 
find  the  locus  of  the  middle  point  of  BC. 

178.  Through  A,  one  of  the  points  of  intersection  of  two  given  circles, 
any  secant  is  drawn  cutting  the  two  circumferences  in  the  points  B  and  C, 
and  on  this  secant  AP  is  laid  off  equal  to  the  sum  of  AB  and  AC]  find  the 
locus  of  P. 

179.  From  a  given  point  0,  any  straight  line   OA  is  drawn  to  a  given 
straight  line  MN,  and  divided  at  P  (internally  or  externally)  so  that  the 
product  OA.  OP  is  equal  to  a  given  constant;  find  the  locus  of  P.  (Ex.  145.) 

180.  From  a  given  point  0  in  the  circumference  of  a  given  circle,  any 
chord  OA  is  drawn  and  divided  at  P  (internally  or  externally)  so  that  the 
product  OA.  OP  is  equal  to  a  given  constant ;  find  the  locus  of  P, 


EXERCISES.  315 


181.  From  a  given  point  0,  any  straight  line  OA  is 
drawn  to  a  given  straight  line  MN,  and  OP  is  drawn 
making  a  given  angle  with  OA,  and  such  that  OP  is  to 
0  A  in  a  given  ratio ;  find  the  locus  of  P. 

With  the  same  construction,  if  OP  is  so  taken  that  the 
product  OP.  OA  is  equal  to  a  given  constant ;  find  the 
locus  of  P. 


182.  From  a  given  point  0,  any  straight  line 
OA  is  drawn  to  a  given  circumference,  and  OP 
is  drawn  making  a  given  angle  with   OA,  and 
such  that  OP  is  to  0 A  in  a  given  ratio ;  find 
the  locus  of  P. 

With  the  same  construction,  if  OP  is  so  taken 
that  the  product  OR  OA  is  equal  to  a  given  constant ;  find  the  locus  of  P. 

183.  One  vertex  of  a  triangle  whose  angles  are  given  is  fixed,  while  the 
second  vertex  moves  on  the  circumference  of  a  given  circle ;  what  is  the 
locus  of  the  third  vertex  ? 

184.  Given  a  circle  0  and  a  point  A  ;  find  the  locus  of  the  point  P  such 
that  the  distance  PA  is  equal  to  the  tangent  from  P  to  the  circle  0. 

185.  Find  the  locus  of  a  point  from  which  two  given  circles  are  seen  under 
equal  angles. 

Note.  The  angle  under  which  a  circle  is  seen  from  a  point  is  the  angle 
contained  by  the  two  tangents  from  that  point. 

186.  Find  the  locus  of  a  point,  such  that  the  sum  of  the  squares  of  its  dis 
tances  from  the  vertices  of  a  given  triangle  is  equal  to  the  square  of  a  given 
line.  (Ex.  166.) 

187.  From  any  point  A  within  a  given  circle,  two  straight  lines  AM  and 
J.JV~are  drawn  perpendicular  to  each  other,  intersecting  the  circumference  in 
M  and  N;  find  the  locus  of  the  middle  point,  of  the  chord  MN. 

PEOBLEMS. 

188.  To  divide  a  given  straight  line  into  three  segments,  A,  .Band  (7,  such 
that  A  and  B  shall  be  in  the  ratio  of  two  given  straight  lines  M  and  'N,  and 
B  and  C  shall  be  in  the  ratio  of  two  other  given  straight  lines  P  and  Q. 

189.  Through  a  given  point,  to  draw  a  straight  line  so  that  the  portion  of 
it  intercepted  between  two  given  straight  lines  shall  be  divided  at  the  point 
in  a  given  ratio. 

190.  Through  a  given  point,  to  draw  a  straight  line  so  that  the  distances 
from  two  other  given  points  to  this  line  shall  be  in  a  given  ratio.  (Two  solu 
tions.  ) 

191.  Through  a  given  point  P,  to  draw  a  straight  line  cutting  two  given 
parallels  in  M  and  N,  so  that  the  distances  AM  and  BN  of  the  points  of 
intersection  from  two  given  points  A  and  B  on  these  parallels  shall  be  in  a 
given  ratio. 


316 


EXERCISES. 


192.  To  determine  a  point  whose  distances  from  three  given  points  shall  be 
proportional  to  three  given  straight  lines.  (III.  79.) 

193.  To  determine  a  point  whose  distances  from  three  given  indefinite 
straight  lines  shall  be  proportional  to  three  given  straight  lines.  (Ex.  170.) 

194.  Given  two  straight  lines  which  cannot  be  produced  to  their  intersec 
tion,  to  draw  a  straight  line  through  a  given  point  which  would,  if  sufficiently 
produced,  pass  through  the  unknown  point  of  intersection  of  the  given  lines. 
(III.  35.) 

195.  In  a  given  triangle  ABC  to  draw  a  parallel  EF  to  the  base  BC, 
intersecting  the  sides  AB  and  AC  in  E  and  F  respectively,   so  that 
BE  +  CF=  BC;  or  so  that  BE—  CF=  BC.  (III.  19,  21.) 


196.  In  a  given  triangle  ABC,  to  inscribe  a  square 
DEFG.  (Exs.  71  and  133.) 


197.  In  a  given  triangle  ABC,  to  inscribe  a  paral 
lelogram  DEFG,  such  that  the  adjacent  sides  DE 
and  D  G  shall  be  in  a  given  ratio  and  contain  a  given 
angle.    (Remark,  that  the  solution  of  this  problem 
includes  that  of  the  preceding. ) 

198.  Construct  a  triangle,  given  its  base,  the  ratio  of  the  other  two  sides, 
and  one  angle.  (III.  79. ) 

199.  To  determine  a  point  in  a  given  arc  of  a  circle,  such  that  the  chords 
drawn  from  it  to  the  extremities  of  the  arc  shall  have  a  given  ratio. 

200.  To  find  a  point  P  in  the  prolongation  of  a  given  chord  CD  of  a 
given  circle,  such  that  the  sum  of  the  two  tangents  PA  and  PB,  drawn  from 
it  to  the  circle,  shall  be  equal  to  the  entire  secant  PC. 

201.  To  divide  a  given  straight  line  into  two  segments,  such  that  the  sum 
of  their  squares  shall  be  equal  to  the  square  of  a  given  straight  line. 

202.  Given  an  obtuse-angled  triangle ;  it  is  required  to  draw  a  straight  line 
from  the  vertex  of  the  obtuse  angle  to  the  opposite  side,  the  square  of  which 
shall  be  equal  to  the  product  of  the  segments  into  which  it  divides  that 
side. 

203.  Through  a  given  point  P  to  draw  a  straight  line  intersecting  a  given 
circumference  in  two  points  A  and  B,  so  that  PA  shall  be  to  PB  in  a  given 
ratio. 


204.  Given  two  circumferences  intersecting 
in  A ;  to  draw  through  A  a  secant,  BA  C,  such 
that  AB  shall  be  to  A  C  in  a  given  ratio. 


EXERCISES. 


317 


205.  Given  two  circumferences  intersecting  in  A ; 
to  draw  through  A  a  secant  ABC,  such  that  the 
product  AB.  AC  shall  be  equal  to  the  square  of  a 
given  line. 

Construction.  Produce  the  common  chord  AD, 
and  take  E  so  that  AD.  AE  =  the  square  of  the 
given  line  (III.  71).  Make  the  angle  AEC  equal 
to  the  angle  inscribed  in  the  segment  ABD,  and 
let  EC  cut  the  circumference  in  C  and  C'  Join 
A  C  and  A  C'.  Either  of  these  lines  satisfies  the 
conditions  of  the  problem. 

206.  To   describe  a  circumference  passing 
through  two  given  points  A  and  B,  and  tan 
gent  to  a  given  circumference  0. 

Analysis.  Suppose  ATB  is  the  required  cir 
cumference  tangent  to  the  given  circumference      / 
at  T,  and  ACDB  any  circumference  passing      | 
through  A  and  B  and  cutting  the  given  cir 
cumference  in  C  and  D.    The  common  chords 
AB  and  CD,  and  the  common  tangent  at  T, 
all  pass  through  a  common  point  P  (Ex.  158) ; 
from  which  a  simple  construction  may  be  inferred.   There  are  two  solutions, 
given  by  the  two  tangents  that  can  be  drawn  from  P. 

207.  To  describe  a  circumference  passing  through  two  given  points  and 
tangent  to  a  given  straight  line.   (Two  solutions. ) 

208.  To  describe  a  circumference  passing  through  a  given  point  and  tan 
gent  to  two  given  straight  lines. 

209.  To    describe  a    circumference 
passing  through  a  given  point  jP,  and 
tangent  to  a  given  straight  line  MN 
and  to  a  given  circumference  0. 

Analysis.  Suppose  EPD  is  one  of 
the  circumferences  which  satisfy  the 
conditions,  passing  through  P,  touch 
ing  MN  at  E  and  the  circumference  0 
at  D.  Through  the  centre  0,  draw 
COB  A  perpendicular  to  MN;  join 
GP  meeting  the  circumference  EPD 
in  Q ;  also  join  CE.  It  can  be  proved 
that  CE  passes  through  D,  and  that 

CPCQ=  CE.CD  =  CA.CB; 

the  point  Q  is  therefore  determined,  and  the  problem  is  reduced  to  that  of 
Ex.  206  or  207.     The  point  Q  may  be  taken  either  in  PC  or  in  PC  pro 
duced  through  C,  and  thus  there  will  be  obtained,  in  all,  four  solutions. 
27* 


318  EXEKCISES. 

210.  To  describe  a  circumfer 
ence   passing   through    a    given 
point  A  and  tangent  to  two  given 
circumferences,  0  and  0/. 

Analysis.  If  ACDB  is  one  of 
the  circumferences  satisfying  the 
conditions,  we  can  show  that  the 
line  CD,  joining  the  points  of  con 
tact  with  the  given  circles,  passes 
through  jP,  the  intersection  of  the 
line  of  centres,  0  0',  with  a  com 
mon  tangent  TT\  and  that 
PC.PD  =  PT.PT'.  Hence,  joining  PA,  we  have  PA.PB  =  PT.PT', 
and  PB  is  known ;  or  B  is  a  known  point  on  the  required  circumference. 
The  problem  is  thus  reduced  to  Ex.  206.  By  employing  also  the  internal 
common  tangent,  we  find,  in  all,  four  solutions. 

211.  To  describe  a  circle  tangent  to  two  given  straight  lines  and  to  a  given 
circle. 

This  is  reducible  to  Ex.  208.  If  both  the  given  straight  lines  intersect  the 
given  circle,  there  may  be  eight  solutions. 

212.  To  describe  a  circle  tangent  to  two  given  circles,  and  to  a  given 
straight  line. 

This  is  reducible  to  Ex.  209.     There  may  be  eight  solutions. 

213.  To  describe  a  circle  tangent  to  three  given  circles. 
This  is  reducible  to  Ex.  210.     There  may  be  eight  solutions. 

*214.  To  describe  a  circumference  which  shall  bisect  three  given  circum 
ferences. 

*215.  To  construct  a  triangle,  given  its  base  in  position  and  magnitude, 
and  the  sum  (or  the  difference)  of  the  other  two  sides,  the  locus  of  the  vertex 
opposite  the  base  being  a  given  straight  line. 

*216.  To  construct  a  triangle,  given  the  product  of  twof  sides,  the  medial 
line  to  the  third  side,  and  the  difference  of  the  angles  adjacent  to  the  third 
side. 

*217.  To  construct  a  triangle,  similar  to  a  given  triangle,  and  having  its 
three  vertices  on  the  circumferences  of  three  given  concentric  circles. 

The  same  problem,  substituting  three  parallel  straight  lines  for  the  three 
circumferences. 

*218.  In  a  given  circle,  to  inscribe  a  triangle,  such  that 

1st.  Its  base  shall  be  parallel  to  a  given  straight  line,  and  its  other  two 
sides  shall  pass  through  two  given  points  in  that  line  ;  or, 

2d.  Its  base  shall  be  parallel  to  a  given  straight  line,  and  its  other  two 
sides  shall  pass  through  two  given  points  not  in  that  line  ;  or, 

3d".  Its  three  sides  shall  pass  through  three  given  points. 

*  Exercises  214  to  218  are  intended  only  for  the  most  advanced  students. 


EXERCISES.  319 

GEOMETRY.— BOOK  IV. 

THEOKEMS. 

219.  Two  triangles  which  have  an  angle  of  the  one  equal  to  the  supple 
ment  of  an  angle  of  the  other  are  to  each  other  as  the  products  of  the  sides 
including  the  supplementary  angles.  (IV.  22.) 

220.  Prove,  geometrically,  that  the  square  described  upon  the  sum  of  two 
straight  lines  is  equivalent  to  the  sum  of  the  squares  described  on  the  two 
lines  plus  twice  their  rectangle. 

Note.  By  the  "rectangle  of  two  lines"  is  here  meant  the  rectangle  of 
which  the  two  lines  are  the  adjacent  sides. 

221.  Prove,  geometrically,  that  the  square  described  upon  the  difference 
of  two  straight  lines  is  equivalent  to  the  sum  of  the  squares  described  on  the 
two  lines  minus  twice  their  rectangle. 

222.  Prove,  geometrically,  that  the  rectangle  of  the  sum  and  the  differ 
ence  of  two  straight  lines  is  equivalent  to  the  difference  of  the  squares  of 
those  lines. 

223.  Prove,  geometrically,  that  the  sum  of  the  squares  on  two  lines  is 
equivalent  to  twice  the  square  on  half  their  sum  plus  twice  the  square  on 
half  their  difference. 

Or,  the  sum  of  the  squares  on  the  two  segments  of  a  line  is  equivalent  to 
twice  the  square  on  half  the  line  plus  twice  the  square  on  the  distance  of  the 
point  of  section  from  the  middle  of  the  line. 

224.  The  area  of  a  triangle  is  equal  to  the  product  of  half  its  perimeter  by 
the  radius  of  the  inscribed  circle  ;  that  is,  if  a,  6  and  c  denote  the  sides  op 
posite  the  angles  A,  B  and  C  respectively,  r  the  radius  of  the  inscribed 
circle,  S  the  area,  and 

s  =  semi-perimeter  =  J  (a  +  b  +  c), 
•  then 

S  =  sr. 

Also,  if  r',  r",  r/x/,  denote  the  radii  of  the  three  escribed  circles,  prove, 
by  Ex.  48  with  the  figure  of  (II.  95),  that 

r        s  —  b  ,,       t  \  , 

__  =  ___,  rr//  =  (s  _  a)  (s  _  c)) 

and  hence  the  following  expressions  for  S,  r,  r',  r",  r'", 


=  V8(s  —  a)  (s  —  b}  (s  — c), 

„///  _    s 


s  —  c 
Also  prove  that 


320 


EXERCISES. 


225.  The  area  of  a  triangle  is  equal  to  the  product  of  its  three  sides 
divided  by  four  times  the  radius  of  the  circumscribed  circle ;  that  is,  de 
noting  this  radius  by  R, 

„_  abc 

*~4R' 
(IV.  13)  and  (III.  65.) 

226.  The  area  of  a  triangle  is  equal  to  the  product  of  the  radius  of  the 
circumscribed  circle  by  the  semi-perimeter  of  the  triangle  formed  by  joining 
the  feet  of  the  perpendiculars  drawn  from  the  vertices  of  the  given  triangle 
to  the  opposite  sides. 

227.  The  area  of  the  triangle  formed  with  the  three  medial  lines  of  a 
given  triangle  is  three-fourths  of  the  area  of  that  triangle.  (IV.  20)  and 
(Ex.  17.) 

228.  The  two  opposite  triangles,  formed  by  joining  any  point  in  the  interior 
of  a  parallelogram  to  its  four  vertices,  are  together  equivalent  to  one-half  the 
parallelogram. 

229.  The  triangle  formed  by  joining  the  middle  point  of  one  of  the  non- 
parallel  sides  of  a  trapezoid  to  the  extremities  of  the  opposite  side,  is  equiva 
lent  to  one-half  the  trapezoid. 

230.  The  figure  formed  by  joining  consecutively  the  four  middle  points  of 
the  sides  of  any  quadrilateral  is  equivalent  to  one-half  the  quadrilateral. 

231.  If  through  the  middle  point  of  each  diagonal  of  any  quadrilateral  a 
parallel  is  drawn  to  the  other  diagonal,  and  from  the  intersection  of  these 
parallels  straight  lines  are  drawn  to  the  middle  points  of  the  four  sides,  these 
straight  lines  divide  the  quadrilateral  into  four  equivalent  parts. 

232.  Two  quadrilaterals  are  equivalent  if  their  diagonals  are  equal,  each 
to  each,  and  contain  equal  angles. 

233.  If  in  a  rectangle  ABCD  we  draw  the 
diagonal  AC,  inscribe  a  circle  in  the  triangle 
ABC,  and  from  its  centre  draw  OE  and  OF 
perpendicular  'to  AD  and  DC,  respectively,  the 
rectangle  OD  will  be  equivalent  to  one-half  the 
rectangle  ABCD. 

234.  Let  ABC  be  any  triangle,  and  upon 
the  sides  AB,  AC,  construct  parallelograms 
AD,  AF,  of  any  magnitude  or  form.     Let 
their  exterior  sides  DE,  FG  meet  in  M ;  join 
MA,  and  upon  BC  construct  a  parallelogram 
BK,  whose,  side  BH  is  equal  and  parallel  to 
MA.     Then  the  parallelogram  BK  is  equiva 
lent  to  the  sum  of  the  parallelograms  AD 
and  AF. 

From  this,  deduce  the  Pythagorean  Theo 
rem.   (IV.  25.) 


EXERCIS  ES. 


321 


*  235.  Upon  the  sides  of  any  triangle 
ABC,  construct  squares  AD,  AG,  BH; 
join  EF,  GH,  DK;  from  A  draw  AP 
perpendicular  to  BC,  and  produce  it  to  Q, 
making  AQ  =  BC]  join  BQ,  CQ,  BG, 
CD,  and  from  D  and  G,  draw  DM,  GN, 
perpendicular  to  BC.  Prove  the  following 
properties : 

1st.  The  triangles  AEF,  CGH,  DKB, 
are  each  equivalent  to  ABC. 

2d.  DM+  GN  =  BC. 

3d.  BQ  is  perpendicular  to  CD,  and 
CQ  to  BG. 

4th.  CD  and  B  G  intersect  on  the  per 
pendicular  AP. 

5th.  The  lines  AQ  and  EF  bisect  each 
other  at  R. 

6th.  EF,  GH,  DK,  are  respectively 
equal  to  twice  the  medial  lines  of  the  triangle  AB  C. 

236.  If  three  straight  lines  Aa,  Bb,  Cc,  drawn 
from  the  vertices  of  a  triangle  ABC  to  the  opposite 
sides,  pass  through  a  common  point  0  within  the 
triangle,  then 

Oa    ,    Ob    ,    Oc  =  -, 

Aa  "*"  Bb       Cc 


H 


What  modification  of  this  statement  is  necessary  if  the  point  0  is  with 
out  the  triangle  ? 

237.  If  from  any  point  0  within  a  triangle 
ABC,  any  three  straight  lines,  Oa,  Ob,  Oc,  are 
drawn  to  the  three  sides,  and  through  the  vertices 
of  the  triangle  three  straight  lines  Aa',  Bb',  Cc', 
are  drawn  parallel  respectively  to  Oa,  Ob,  Oc, 
then 

Oa     .Ob     ,    J9c  ^  -. 

Aa'  "•"  Bb'  "*"  Cc' 


What  modification  of  this  statement  is  necessary  if  the  point  0  is  taken 
without  the  triangle  ? 
Deduce  the  preceding  theorem  from  this. 


238.  If  from  the  vertices  of  a  triangle  ABC,  three  straight  lines,  A  A', 
BB',  CC',  are  drawn  to  the  opposite  sides  (or  these  sides  produced),  each 
equal  to  a  given  line  L,  and  from  any  point  0  within  the  triangle,  Oa,  Ob, 

V 


322  EXERCISES. 

0c,  are  drawn  parallel  respectively  to  A  A',  BB',  CC',  and  terminating 
in  the  same  sides,  then,  the  sum  of  0a,  Ob  and  Oc  is  equal  to  the  given 
length  L. 


239.  If  a,  £>,  c  ana  d  denote  the  four  sides  of  any  quadrilateral,  m  and  n 
its  diagonals,  and  S  its  area,  then  .    . 


n  +  oa  —  ft8  +  c2  —  d2}  (2mn  —  a*  +  b*  —  c2  +  d2}. 
If  the  quadrilateral  is  inscribed  in  a  circle,  this  formula  becomes 


—  a}(p  —  b)  (p  —  c)(p  —  d), 

in  which  p  =  \  (a  +  6  +  c  +  d). 

If  the  quadrilateral  is  such  that  it  can  be  circumscribed  about  a  circle  and 
also  inscribed  in  a  circle,  then  the  formula  becomes 


PROBLEMS. 

240.  To  construct  a  triangle,  given  one  angle,  the  side  opposite  to  that 
angle,  and  the  area  (equal  to  that  of  a  given  square).     v, 

241.  To  construct  a  triangle,  given  its  angles  and  its  area. 

242.  To  construct  a  triangle,  given  one  angle,  the  medial  line  from  one  of 
the  other  angles,  and  the  area. 

243.  To  construct  a  triangle,  given  its  area,  the  radius  of  the  inscribed 
circle,  and  the  radius  of  one  of  the  escribed  circles  ;  or,  given  its  area  and 
the  radii  of  two  escribed  circles.  (Exercises  48  and  224.  ) 

244.  Given  any  triangle,  to  construct  an  isosceles  triangle  of  the  same 
area,  whose  vertical  angle  is  an  angle  of  the  given  triangle. 

245.  Given  any  triangle,  to  construct  an  equilateral  triangle  of  the  same 
area. 

246.  Given  the  three  straight  lines  EF,  GH  and  DK,  in  the  figure  of 
Exercise  235,  to  construct  the  triangle  AB  C. 

247.  Bisect  a  given  triangle  by  a  parallel  to  one  of  its  sides. 

Or,  more  generally,  divide  a  given  triangle  into  two  or  more  parts  propor 
tional  to  given  lines,  by  parallels  to  one  of  its  sides. 


EXERCISES.  323 

248.  Bisect  a  triangle  by  a  straight  line  drawn  through  a  given  point  in 
one  of  its  sides. 

249.  Through  a  given  point,  draw  a  straight  line  which  shall  form  with 
two  given  intersecting  straight  lines  a  triangle  of  a  given  area. 

Remark  that  the  area  and  an  angle  being  known,  the  product  of  the  sides 
including  that  angle  is  known.   (IV.  22. ) 

250.  Bisect  a  trapezoid  by  a  parallel  to  its  bases. 

251.  Inscribe  a  rectangle  of  a  given  area  in  a  given  circle. 

252.  Inscribe  a  trapezoid  in  a  given  circle,  knowing  its  area  and  the 
common  length  of  its  inclined  sides.   (See  Ex.  229.) 

253.  Given  three  points,  J.,  B  and  (7,  to  find  a  fourth  point  P,  such  that 
the  areas  of  the  triangles  APE,  APC,  BPC,  shall  be  equal.    (Four  solu 
tions.  ) 

254.  Given  three  points,  A,  B  and  (7,  to  find  a  fourth  point  P,  such  that 
the  areas  of  the  triangles  APB,  APC,  BPC,  shall  be  proportional  to  three 
given  lines  L,  M,  N.    (Four  solutions. ) 

See  Exercise  170. 


GEOMETRY.— BOOK  V. 

THEOREMS. 

255.  An  equilateral  polygon  inscribed  in  a  circle  is  regular. 

256.  An  equilateral  polygon  circumscribed  about  a  circle  is  regular  if  the 
number  of  its  sides  is  odd. 

257.  An  equiangular  polygon  inscribed  in  a  circle  is  regular  if  the  number 
of  its  sides  is  odd. 

258.  An  equiangular  polygon  circumscribed  about  a  circle  is  regular. 

259.  The  area  of  the  regular  inscribed  hexagon  is  three-fourths  of  that 
of  the  regular  circumscribed  hexagon. 

260.  The  area  of  the  regular  inscribed  hexagon  is  a  mean  proportional 
between  the  areas  of  the  inscribed  and  circumscribed  equilateral  triangles. 

261.  A  plane  surface  may  be  entirely  covered  (as  in  the  construction  of  a 
pavement)  by  equal  regular  polygons  of  either  three,  four,  or  six  sides. 

262.  A  plane  surface  may  be  entirely  covered  by  a  combination  of  squares 
and  regular  octagons  having  the  same  side,  or  by  dodecagons  and  equilateral 
triangles  having  the  same  side. 

263.  The  area  of  a  regular  inscribed  octagon  is  equal  to  that  of  a  rectangle 
whose  adjacent  sides  are  equal  to  the  sides  of  the  inscribed  and  circumscribed 
squares. 

264.  The  area  of  a  circle  is  a  mean  proportional  between  the  areas  of  any 
two  similar  polygons,  one  of  which  is  circumscribed  about  the  circle  and  the 
other  isoperinietrical  with  the  circle.  (Galileo's  Theorem.} 


324 


EXERCISES. 


265.  Two  diagonals  of  a  regular  pentagon,  not  drawn  from  a  common 
vertex,  divide  each  other  in  extreme  and  mean  ratio. 

266.  If  a  =  the  side  of  a  regular  pentagon  inscribed  in  a  circle  whose 
V    radius  is  R,  then, 

a  =  f  1/10  —  2i/5. 

267.  If  a  =  the  side  of  a  regular  octagon  inscribed  in  a  circle  whose  radius 
is  R,  then,  _ 

a  =  RV2  —  1/2. 

268.  If  a  =  the  side  of  a  regular  dodecagon  inscribed  in  a  circle  whose 
radius  is  J?,  then,  _ 

a  =  RV2  —  i/3. 

269.  If  a  =  the  side  of  a  regular  pentedecagon  inscribed  in  a  circle  whose 
radius  is  R,  then, 


+  2i/5  +  1/3  —  1/15). 


270.  If  <Z  =  the  diagonal  of  a  regular  pentagon  inscribed  in  a  circle  whose 
radius  is  -K,  then, 

d  ='-  £  1/10  +  2i/5. 

271.  If  a  =  the  side  of  a  regular  polygon  inscribed  in  a  circle  whose  radius 
is  R,  and  A  =  the  side  of  the  similar  circumscribed  polygon,  then, 


2aR 


2AR 


-a2) 

272.  If  a  =  the  side  of  a  regular  polygon  inscribed  in  a  circle  whose  radius 
is  RJ  and  a'  =  the  side  of  the  regular  inscribed  polygon  of  double  the 
number  of  sides,  then, 


273.  If  AB  and  CD  are  two  perpendicular  di 
ameters  in  a  circle,  and  E  the  middle  point  of  the 
radius  OC,  and  if  EF  is  taken  equal  to  EA,  then 
OF  is  equal  to  the  side  of  the  regular  inscribed  A 
decagon,  and  AF  is  equal  to  the  side  of  the  regular 
inscribed  pentagon. 

Corollary.  If  a  =  the  side  of  a  regular  pentagon 
and  a'  =  the  side  of  a  regular  decagon,  inscribed 
in  a  circle  whose  radius  is  R,  then, 


EXEKCISES.  325 

274.  In  two  circles  of  different  radii,  angles  at  the  centres  subtended  by 
arcs  of  equal  length  are  to  each  other  inversely  as  the  radii. 

275.  From  any  point  within  a  regular  polygon  of  n  sides,  perpendiculars 
are  drawn  to  the  several  sides  ;  prove  that  the  sum  of  these  perpendiculars 
is  equal  to  n  times  the  apothem.  (V.  22.  ) 

What  modification  of  this  statement  is  required  if  the  point  is  taken  with 
out  the  polygon? 

276.  If  perpendiculars  are  dropped  from  the  vertices  of  a  regular  polygon 
upon  any  diameter  of  the  circumscribed  circle,  the  sum.  of  the  perpendicu 
lars  which  fall  on  one  side  of  this  diameter  is  equal  to  the  sum  of  those  which 
fall  on  the  opposite  side. 

277.  If  n  is  the  number  of  sides  of  a  regular  polygon  inscribed  in  a  circle 
whose  radius  is  R,  and  a  point  P  is  taken  such  that  the  sum  of  the  squares 
of  its  distances  from  the  vertices  of  the  polygon  is  equal  to  a  given  quantity 
&2,  the  locus  of  P  is  the  circumference  of  a  circle,  concentric  with  the 
given  circle,  whose  radius  r  is  determined  by  the  relation 


(III.  52  and  53),  (Ex.  276.) 

PROBLEMS. 

278.  Divide  a  given  circle  into  a  given  number  of  equivalent  parts,  by  con 
centric  circumferences. 

Also,  divide  it  into  a  given  number  of  parts  proportional  to  given  lines,  by 
concentric  circumferences. 

279.  A  circle  being  given,  to  find  a  given  number  of  circles  whose  radii 
shall  be  proportional  to  given  lines,  and  the  sum  of  whose  areas  shall  be 
equal  to  the  area  of  the  given  circle. 

280.  In  a  given  equilateral  triangle,  inscribe  three  equal  circles  tangent  to 
each  other  and  to  the  sides  of  the  triangle. 

Determine  the  radius  of  these  circles  in  terms  of  the  side  of  the  triangle. 

281.  In  a  given  circle,  inscribe  three  equal  circles  tangent  to  each  other 
and  to  the  given  circle. 

Determine  the  radius  of  these  circles  in  terms  of  the  radius  of  the  given 
circle. 


GEOMETRY.— BOOK   VI. 

THEOREMS. 

282.  If  a  straight  line  AB  is  parallel  to  a  plane  MN,  any  plane  perpen 
dicular  to  the  line  AB  is  perpendicular  to  the  plane  MN. 

283.  If  a  plane  is  passed  through  one  of  the  diagonals  of  a  parallelogram, 
the  perpendiculars  to  this  plane  from  the  extremities  of  the  other  diagonal 
are  equal. 

28 


326  EXERCISES. 

284.  If  the  intersections  of  a  number  of  planes  are  parallel,  all  the  per 
pendiculars  to  these  planes,  drawn  from  a  common  point  in  space,  lie  in  one 
plane. 

285.  If  the  projections  of  a  number  of  points  on  a  plane  are  in  a  straight 
line,  these  points  are  in  one  plane. 

286.  If  each  of  the  projections  of  a  line  AB  upon  two  intersecting  planes 
is  a  straight  line,  the  line  AB  is  a  straight  line. 

287.  Let  A  and  B  be  two  points,  and  M  and  N  two  planes.     If  the  sum 
of  the  two  perpendiculars  from  the  point  A  upon  the  planes  M  and  N  is 
equal  to  the  sum  of  those  from  B  upon  these  planes,  this  sum  is  the  same 
for  every  other  point  in  the  straight  line  AB.  (Ex.  125.) 

Extend  the  theorem  to  any  number  of  planes. 

288.  Let  J.,  B  and  C  be  three  points,  and  M  and  N  two  planes.     If  the 
sum  of  the  two  perpendiculars  from  each  of  the  points  A,  B  and  (7,  upon 
the  planes  M  and  N,  is  the  same  for  the  three  points,  it  will  be  the  same 
for  every  point  in  the  plane  ABC.  (Ex.  287.) 

Extend  the  theorem  to  any  number  of  planes. 

289.  A  plane  passed  through  the  middle  point  of  the  common  perpen 
dicular  to  two  straight  lines  in  space  (VI.  63),  and  parallel  to  both  these 
lines,  bisects  every  straight  line  joining  a  point  of  one  of  these  lines  to  a 
point  of  the  other. 

290.  In  any  triedral  angle,  the  three  planes  bisecting  the  three  diedral 
angles,  intersect  in  the  same  straight  line. 

291.  In  any  triedral  angle,  the  three  planes  passed  through  the  edges,  per 
pendicular  to  the  opposite  faces  respectively,  intersect  in  the  same  straight 
line. 

292.  In  any  triedral  angle,  the  three  planes  passed  through  the  edges  and 
the  bisectors  of  the  opposite  face  angles  respectively,  intersect  in  the  same 
straight  line. 

293.  In  any  triedral  angle,  the  three  planes  passed  through  the  bisectors 
of  the  face  angles,  and  perpendicular  to  these  faces  respectively,  intersect  in 
the  same  straight  line. 

294.  If  through  the  vertex  of  a  triedral  angle,  three'  straight  lines  are 
drawn,  one  in  the  plane  of  each  face  and  perpendicular  to  the  opposite  edge, 
these  three  straight  lines  are  in  one  plane. 


LOCI. 

295.  Find  the  locus  of  the  points  in  space  which  are  equally  distant  from 
two  given  points. 

296.  Locus  of  the  points  which  are  equally  distant  from  two  given  planes ; 
or  whose  distances  from  two  given  planes  are  in  a  given  ratio.  (Compare 
Ex.  170.) 

297.  Locus  of  the  points  which  are  equally  distant  from  two  given  straight 
lines  in  the  same  plane. 

298.  Locus  of  the  points  which  are  equally  distant  from  three  given 
points. 


EXERCISES.  327 

299.  Locus  of  the  points  which  are  equally  distant  from  three  given 
planes. 

300.  Locus  of  the  points  which  are  equally  distant  from  three  given 
straight  lines  in  the  same  plane. 

301.  Locus  of  the  points  which  are  equally  distant  from  the  three  edges 
of  a  triedral  angle.  (Ex.  293.) 

302.  Locus  of  the  points  in  a  given  plane  which  are  equally  distant  from 
two  given  points  out  of  the  plane. 

303.  Locus  of  the  points  which  are  equally  distant  from  two  given  planes, 
and  at  the  same  time  equally  distant  from  two  given  points.  (Exs.  295  and 
296.) 

304.  Locus  of  a  point  in  a  given  plane  such  that  the  straight  lines  drawn 
from  it  to  two  given  points  out  of  the  plane  make  equal  angles  with  the 
plane.    (III.  79.) 

305.  Locus  of  a  point  such  that  the  sum  of  its  distances  from  two  given 
planes  is  equal  to  a  given  straight  line. 

306.  Locus  of  a  point  such  that  the  difference  of  the  squares  of  its  dis 
tances  from  two  given  points  is  equal  to  a  given  constant. 

307.  Locus  of  a  point  in  a  given  plane  such  that  the  difference  of  the 
squares  of  its  distances  from  two  given  points  is  equal  to  a  given  constant. 

308.  A  straight  line  of  a  given  length  moves  so  that  its  extremities  are 
constantly  upon  two  given  perpendicular  but  non-intersecting  straight  lines ; 
what  is  the  locus  of  the  middle  point  of  the  moving  line  ? 

309.  Two  given  non-intersecting  straight  lines  in  space  are  cut  by  an 
indefinite  number  of  parallel  planes,  the  two  intersections  of  each  plane 
with  the  given  lines  are  joined  by  a  straight  line,  and  each  of  these  joining 
lines  is  divided  in  a  given  ratio  m  :  n ;  what  is  the  locus  of  the  points  of 
division  ? 

PEOBLEMS. 

In  the  solution  of  problems  in  space,  we  assume — 1st,  that  a  plane  can  be 
drawn  passing  through  three  given  points  (or  two  intersecting  straight  lines) 
and  its  intersections  with  given  straight  lines  or  planes  determined— and  2d, 
that  a  perpendicular  to  a  given  plane  can  be  drawn  at  a  given  point  in  the 
plane,  or  from  a  given  point  without  it ;  and  the  solution  of  a  problem  will 
consist,  not  in  giving  a  graphic  construction,  but  in  determining  the  con 
ditions  under  which  the  proposed  problem  is  solved  by  the  application  of 
these  elementary  problems.  The  graphic  solution  of  problems  belongs  to 
Descriptive  Geometry. 

310.  Through  a  given  straight  line,  to  pass  a  plane  perpendicular  to  a 
given  plane. 

311.  Through  a  given  point,  to  pass  a  plane  perpendicular  to  a  given 
straight  line. 

312.  Through  a  given  point,  to  pass  a  plane  parallel  to  a  given  plane. 

313.  To  determine  that  point  in  a  given  straight  line  which  is  equidistant 
from  two  given  points  not  in  the  same  plane  with  the  given  line. 


328  EXERCISES. 

314.  To  find  a  point  in  a  plane  which  shall  be  equidistant  from  three  given 
points  in  space. 

315.  Through  a  given  point  in  space,  to  draw  a  straight  line  which  shall 
cut  two  given  straight  lines  not  in  the  same  plane. 

316.  Given  a  straight  line  AB  parallel  to  a  plane  M ;  from  any  point  A 
in  AB,  to  draw  a  straight  line  AP,  of  a  given  length,  to  the  plane  M, 
making  the  angle  BAP  equal  to  a  given  angle. 

317.  Through  a  given  point  A  in  a  plane,  to  draw  a  straight  line  AT  in 
that  plane,  which  shall  be  at  a  given  distance  PT  from  a  given  point  P 
without  the  plane. 

318.  A  given  straight  line  AB  meets  a  given  plane  at  the  point  A ;  to 
draw  through  A  a  straight  line  AP  in  the  given  plane,  making  the  angle 
BAP  equal  to  a  given  angle. 

319.  Through  a  given  point  A,  to  draw  to  a  given  plane  M  a  straight  line 
which  shall  be  parallel  to  a  given  plane  JV  and  of  a  given  length. 

320.  Through  a  given  point  A,  to  draw  to  a  given  plane  M  a  straight  line 
which  shall  be  parallel  to  a  given  plane  N  and  make  a  given  angle  with  the 
plane  M. 

321.  Given  two  straight  lines,  CD  and  EF,  not  in  the  same  plane,  and 
AB  any  third  straight  line  in  space ;  to  draw  a  straight  line  PQ  from  AB  to 
EF  which  shall  be  parallel  to  CD. 

322.  Given  two  straight  lines  AB  and  CD,  not  in  the  same  plane ;  to 
draw  a  straight  line  PQ  from  AB  to  CD  which  shall  make  a  given  angle 
with  AB. 

323.  Given  two  straight  lines,  AB  and  CD,  not  in  the  same  plane,  to  find 
a  point  in  AB  at  a  given  perpendicular  distance  from  CD. 

324.  Through  a  given  point ,  to  draw  a  straight  line  which  shall  meet  a 
given  straight  line  and  the  circumference  of  a  given  circle  not  in  the  same 
plane. 

325.  In  a  given  plane  and  through  a  given  point  of  the  plane,  to  draw 
a  straight  line  which  shall  be  perpendicular  to  a  given   line   in   space. 
(VI.  62.) 

326.  In  a  given  plane,  to  determine  a  point  such  that  the  sum  of  its  dis 
tances  from  two  given  points  on  the  same  side  of  the  plane  shall  be  a 
minimum. 

327.  In  a  given  plane,  to  determine  a  point  such  that  the  difference  of  its 
distances  from  two  given  points  on  opposite  sides  of  the  plane  shall  be  a 
maximum. 

328.  To  cut  a  given  polyedral  angle  of  four  faces  by  a  plane  so  that  the 
section  shall  be  a  parallelogram. 


EXEBCISES.  329 


GEOMETRY.— BOOK  VII. 

THEOKEMS. 

329.  The  volume  of  a  triangular  prism  is  equal  to  the  product  of  the  area 
of  a  lateral  face  by  one-half  the  perpendicular  distance  of  that  face  from  the 
opposite  edge. 

330.  In  any  quadrangular  prism,  the  sum  of  the  squares  of  the  twelve 
edges  is  equal  to  the  sum  of  the  squares  of  its  four  diagonals  plus  eight 
times  the  square  of  the  line  joining  the  common  middle  points  of  the  diago 
nals  taken  two  and  two. 

Deduce  (VII.  20)  from  this. 

331.  Of  all  quadrangular  prisms  having  equivalent  surfaces,  the  cube  has 
the  greatest  volume. 

332.  The  lateral  surface  of  a  pyramid  is  greater  than  the  baee. 

333.  At  any  point  in  the  base  of  a  regular  pyramid  a  perpendicular  to  the 
base  is  erected  which  intersects  the  several  lateral  faces  of  the  pyramid,  or 
these  faces  produced.     Prove  that  the  sum  of  the  distances  of  the  points  of 
intersection  from  the  base  is  constant. 

(See  Ex.  275.) 

334.  In  a  tetraedron,  the  planes  passed  through  the  three  lateral  edges 
and  the  middle  points  of  the  edges  of  the  base  intersect  in  a  straight  line. 
The  four  straight  lines  so  determined,  by  taking  each  face  as  a  base,  meet  in 

1  a  point  which  divides  each  line  in  the  ratio  1  :  4. 

Note.  This  point  is  the  centre  of  gravity  of  the  tetraedron. 

335.  The  perpendicular  from  the  centre  of  gravity  of  a  tetraedron  to  any 
plane  is  equal  to  the  arithmetical  mean  of  the  four  perpendiculars  from  the 
vertices  of  the  tetraedron  to  the  same  plane.  (Ex.  125. }\ 

336.  In  any  tetraedron,  the  straight  lines  joining  the  middle  points  of  the 
opposite  edges  meet  in  a  point  and  bisect  each  other  in  that  point. 

337.  The  plane  which  bisects  a  diedral  angle  of  a  tetraedron  divides  the 
opposite  edge  into  segments  which  are  proportional  to  the  areas  of  the  adja 
cent  faces. 

338.  Any  plane  passing  through  the  middle  points  of  two  opposite  edges 
of  a  tetraedron  divides  the  tetraedron  into  two  equivalent  solids. 

339.  If  one  of  the  triedral  angles  of  a  tetraedron  is   tri-rectangular 
(i.  e.,  composed  of  three  right  angles),  the  square  of  the  area  of  the  face 
opposite  to  it  is  equal  to  the  sum  of  the  squares  of  the  areas  of  the  three 
other  faces. 

340.  If  a,  5,  c,  c?,  are  the  perpendiculars  from  the  vertices  of  a  tetraedron 
upon  the  opposite  faces,  and  a',  fe',  c7,  d',  the  perpendiculars  from  any  point 
within  the  tetraedron  upon  the  same  faces  respectively,  then, 


of    ,    V    ,   c'    ,   d'  _ 

r  7~   i         r   7  —  *• 
abed 


28* 


330  EXERCISES. 

341.  If  ABCD  is  any  tetraedron,  and  0  any  point  within  it;  and  if  the 
straight  lines  AO,  BO,  CO,  DO,  are  produced  to  meet  the  faces  in  the 
points  a,  6,  c,  e?,  respectively ;  then 

Q?L  4-  Qb  4-  0°  4.  Q2  =  1 

4a       £6       6Yc  "*"  Z>d 

342.  The  volume  of  a  truncated  triangular  prism  is  equal  to  the  product 
of  the  area  of  its  lower  base  by  the  perpendicular  upon  the  lower  base  let 
fall  from  the  intersection  of  the  medial  lines  of  the  upper  base. 

343.  The  volume  of  a  truncated  parallelepiped  is  equal  to  the  product  of 
the  area  of  its  lower  base  by  the  perpendicular  from  the  centre  of  the  upper 
base  upon  the  lower  base. 

344.  The  volume  of  a  truncated  parallelepiped  is  equal  to  the  product  of 
a  right  section  by  one-fourth  the  sum  of  its  four  lateral  edges.  (VII.  62. ) 

345.  The  altitude  of  a  regular  tetraedron  is  equal  to  the  sum  of  the  four 
perpendiculars  let  fall  from  any  point  within  it  upon  the  four  faces. 

346.  Any  plane  passed  through  the  centre  of  a  parallelepiped  divides  it 
into  two  equivalent  solids. 


PROBLEMS. 

347.  To  cut  a  cube  by  a  plane  so  that  the  section  shall  be  a  regular 
hexagon. 

348.  Given  three  indefinite  straight  lines  in  space  which  do  not  intersect, " 
to  construct  a  parallelepiped  which  shall  have  three  of  its  edges  on  these 
lines. 

349.  A  parallelepiped  is  given  in  position,  and  a  straight  line  in  space  ;  to 
pass  a  plane  through  the  line  which  shall  divide  the  parallelepiped  into  two 
equivalent  solids. 

350.  To  find  two  straight  lines  in  the  ratio  of  the  volumes  of  two  given 
cubes.  f 

351.  Within  a  given  tetraedron,  to  find  a  point  sucn  that  planes  passed 
through  this  point  and  the  edges  of  the  tetraedron  shall  divide  the  tetraedron 
into  four  equivalent  tetraedrons. 

352.  To  pass  a  plane,  either  through  a  given  point,  or  parallel  to  a  given 
straight  line,  which  shall  divide  a  given  tetraedron  into  two  equivalent 
solids. 

353.  Find  the  difference  between  the  volume  of  the  frustum  of  a  pyramid 
and  the  volume  of  a  prism  of  the  same  altitude  whose  base  is  a  section  of 
the  frustum  parallel  to  its  bases  and  equidistant  from  them. 

The  difference  may  be  expressed  in  the  form  -^V  B  —  V^r,  if  B  and 
6  are  the  areas  of  the  bases,  and  h  the  altitude  of  the  frustum. 


EXEKCISES.  331 


GEOMETRY.— BOOKS  VIII  AND  IX. 

THEOREMS. 

354.  If  through,  a  fixed  point,  within  or  without  a  sphere,  three  straight 
lines  are  drawn  perpendicular  to  each  other,  intersecting  the  surface  of  the 
sphere,  the  sum  of  the  squares  of  the  three  intercepted  chords  is  constant. 
Also,   the  sum  of  the  squares  of  the  six  segments  of  these  chords  is 
constant. 

355.  If  three  radii  of  a  sphere,  perpendicular  to  each  other,  are  projected 
upon  any  plane,  the  sum  of  the  squares  of  the  three  projections  is  equal  to 
twice  the  square  of  the  radius  of  the  sphere.  (Ex.  339.) 

356.  If  two  circles  revolve  about  the  line  joining  their  centres,  a  common 
tangent  to  the  two  circles  generates  the  surface  of  a  common  tangent  cone  to 
the  two  spheres  generated  by  the  circles.     The  vertex  of  the  cone  generated 
by  an  external  common  tangent  may  be  called  an  external  vertex,  and  that 
of  the  cone  generated  by  an  internal  common  tangent  may  be  called  an 
internal  vertex.     These  terms  being  premised,  prove  the  following  theorem : 

If  three  spheres  of  different  radii  are  placed  in  any  position  in  space, 
and  the  six  common  tangent  cones,  external  and  internal,  are  drawn  to  these 
spheres  taken  two  and  two,  1st,  the  three  external  vertices  are  in  a  straight 
line  ;  2d,  each  external  vertex  lies  in  the  same  straight  line  with  two  internal 
vertices. 

357.  The  volumes  of  a  cone  of  revolution,  a  sphere,  and  a  cylinder  of 
revolution,  are  proportional  to  the  numbers  1,  2,  3,  if  the  bases  of  the  cone 
and  cylinder  are  each  equal  to  a  great  circle  of  the  sphere,  and  their  altitudes 
are  each  equal  to  a  diameter  of  the  sphere. 

358.  An  equilateral  cylinder  (of  revolution)  is  one  a  section  of  which 
through  the  axis  is  a  square.     An  equilateral  cone  (of  revolution)  is  one  a 
section  of  which  through  the  axis  is  an  equilateral  triangle.     These  defi 
nitions  premised,  prove  the  following  theorems : 

I.  The  total  area  of  the  equilateral  cylinder  inscribed  in  a  sphere  is  a 
mean  proportional  between  the  area  of  the  sphere  and  the  total  area  of  the 
inscribed  equilateral  cone.     The  same  is  true  of  the  volumes  of  these  three 
bodies. 

II.  The  total  area  of  the  equilateral  cylinder  circumscribed  about  a  sphere 
is  a  mean  proportional  between  the  area  of  the  sphere  and  the  total  area  of 
the  circumscribed  equilateral  cone.     The  same  is  true  of  the  volumes  of 
these  three  bodies. 

359.  If  h  is  the  altitude  of  a  segment  of  one  base  in  a  sphere  whose 
radius  is  r,  the  volume  of  the  segment  is  equal  to  ^h^(R  —  i/i). 

360.  The  volumes  of  polyedrons  circumscribed  about  the  same  sphere  are 
proportional  to  their  surfaces. 


332 


EXERCISES. 


LOCI. 


361.  Locus  of  the  points  in  space  which  are  at  a  given  distance  from  a 
given  straight  line. 

362.  Locus  of  the  points  which  are  at  the  distance  a  from  a  point  A,  and 
at  the  distance  b  from  a  point  B. 

363.  Locus  of  the  centres  of  the  spheres  which  are  tangent  to  three  given 
planes. 

364.  Locus  of  a  point  in  space  the  ratio  of  whose  distances  from  two  fixed 
points  is  a  given  constant. 

365.  Locus  of  the  centres  of  the  sections  of  a  given  sphere  made  by  planes 
passing  through  a  given  straight  line. 

366.  Locus  of  the  centres  of  the  sections  of  a  given  sphere  made  by  planes 
passing  through  a  given  point. 

367.  Locus  of  a  point  in  space  the  sum  of  the  squares  of  whose  distances 
from  two  fixed  points  is  a  given  constant.   (Ex.  174.) 

368.  Locus  of  a  point  in  space  the  difference  of  the  squares  of  whose  dis 
tances  from  two  fixed  points  is  a  given  constant.   (Ex.  175.) 


PEOBLEMS. 

369.  To  cut  a  given  sphere  by  a  plane  passing  through  a  given  straight 
line  so  that  the  section  shall  have  a  given  radius. 

370.  To  construct  a  spherical  surface  with  a  given  radius,  1st,  passing 
through  three  given  points ;  2d,  passing  through  two  given  points  and  tan 
gent  to  a  given  plane,  or  to  a  given  sphere ;   3d,  passing  through  a  given 
point  and  tangent  to  two  given  planes,  or  to  two  given  spheres,  or  to  a  given 
plane  and  a  given  sphere ;  4th,  tangent  to  three  given  planes,  or  to  three 
given  spheres,  or  to  two  given  planes  and  a  given  sphere,  or  to  a  given  plane 
and  two  given  spheres. 

371.  Through  a  given  point  on  the  surface  of  a  sphere,  to  draw  a  great 
circle  tangent  to  a  given  small  circle. 

372.  To  draw  a  great  circle  tangent  to  two  given  small  circles. 

373.  At  a  given  point  in  a  great  circle,  to  draw  an  arc  of  a  great  circle 
which  shall  make  a  given  angle  with  the  first. 

374.  To  find  the  ratio  of  the  volumes  of  two  cylinders  whose  convex  areas 
are  equal. 

375.  To  find  the  ratio  of  the  convex  areas  of  two  cylinders  whose  volumes 
are  equal. 

376.  To  find  the  ratio  of  the  volumes  generated  by  a  rectangle  revolving 
successively  about  its  two  adjacent  sides. 


APPENDIX    II. 


INTRODUCTION  TO  MODERN  GEOMETRY. 


INTRODUCTION  TO  MODERN  GEOMETRY. 


TRANSVERSALS. 

1.  DEFINITION.   Any  straight  line  cutting  a  system  of  lines  is  called  a 
transversal. 

PROPOSITION"   I.— THEOREM. 

2.  If  a  transversal  cuts  the  sides  of  a  triangle  (produced  if  necessary),  the 
product  of  three  non-adjacent  segments  of  the  sides  is  equal  to  the  product 
of  the  other  three  segments. 

Let  ABC  be  the  given  triangle,  A 

and  abc  the  transversal.  When  the 
transversal  cuts  a  side  produced,  as 
the  side  BC  at  a,  the  segments  are 
the  distances,  aB,  aC,  of  the  point 
of  section  from  the  extremities  of 
the  line  (III.  22).  The  segments 
aB,  bC,  cA,  having  no  extremity 
in  common,  are  non-adjacent. 

Draw  CN  parallel  to  AB. 
similar  triangles,  we  have 

a£  =  _cB^  bC__NC 

aC      NC*  bA~  cA' 

and  multiplying, 

aCXbA  =  ~cA' 
whence, 

aB  X  bCX  cA  =  aCX  bA  X  cB. 

3.  Corollary.    Conversely,  if  three  points  are  taken  on  the  sides  of  a  tri 
angle  (one  of  the  points,  or  all  three,  lying  in  the  sides  produced),  so  that 
the  product  of  three  non-adjacent  segments  of  the  sides  is  equal  to  the  product 
of  the  other  three,  then  the  three  points  lie  in  the  same  straight  line. 

^  Let  a,  b,  c,  be  so  taken  on  the  sides  of  the  triangle  ABC,  that  the  rela 
tion  [I]  is  satisfied.      Join  ab,  and  let  ab  produced  be  supposed  to  cut 

335 


336  MODERN     GEOMETRY. 

AB  in  a  point  which  we  shall  call-c'.    Then  by  the  above  theorem,  we 
have 

aBXbCX  cfA  =  aCXbAX 


and  since  by  hypothesis  we  also  have 

aB  X  bCX  cA  =  aC  XbAX  cB, 

there  follows,  by  division, 

SA^^B 
cA       cB' 

which  can  evidently  be  true  only  when  </  coincides  with  c;  that  is,  the 
three  points  a,  b  and  c  are  in  the  same  straight  line. 

4.  Scholium.  The  principle  in  the  corollary  often  serves  to  determine,  in  a 
very  simple  manner,  whether  three  points  lie  in  the  same  straight  line. 

For  example,  take  the  following  theorem  : 

The  middle  points  of  the  three  diagonals  of  a  complete  quadrilateral  are 
in  a  straight  line. 

A  complete  quadrilateral  is  the  figure  formed  by  four  straight  lines  inter 
secting  in  six  points,  as  ABCDEF.  The  line  EF  is  called  the  third 
diagonal. 


Let  L,  Mj  N,  be  the  middle  points  of  the  three  diagonals.  Let  G,  ff,  K, 
be  the  middle  points  of  the  sides  of  the  triangle  FDC.  The  sides  of  the 
triangle  GHK  pass  through  the  points  LMN,  respectively  (1. 121  and  122). 
The  line  ABE,  considered  as  a  transversal  of  the  triangle  CDF,  gives 

AD.  BF.  EC  =  AF.B  C.  ED. 

Dividing  each  factor  of  this  equation  by  2,  and  observing  that  we  have 
\AD  =  LK,  %BF  =  MG,  etc.  (I.  121),  we  find 

LK.MG.NH^  LH.MK.NG\ 

therefore,  the  points  L,  M,  N,  lying  in  the  sides  of  ^the  triangle  GHK, 
satisfy  the  condition  of  the  preceding  corollary,  and  are  in  a  straight  line. 


MODERN      GEOMETRY.  337 


PKOPOSITION    II.— THEOEEM. 

5.  Three  straight  lines,  drawn  through  the  vertices  of  a  triangle  and  any 
point  in  its  plane,  divide  the  sides  into  segments  such  that  the  product  of 
three  non-adjacent  segments  is  equal  to  the  product  of  the  other  three. 


Let  ABC  be  the  triangle,  and  0  any  point 
in  its  plane,  through  which  Aa,  Bb,  Cc,  are 
drawn. 

The  triangle  A  Ca  is  cut  by  the  transversal 
Bb  ;  hence,  by  (2), 


and  the  triangle  ABa,  cut  by  the  transversal 
Cc,  gives  B  c  a 

BC.a0.cA  =  aC.AO.cB. 

Multiplying  these  equations  together,  and  omitting  the  common  factors,  we 
obtain 


6.  Corollary.  Conversely,  if  three  points  are  taken  on  the  sides  of  a  tri 
angle  (all  the  points  being  on  the  sides  themselves  or  two  on  the  sides  pro 
duced),  so  that  the  product  of  three  non-adjacent  segments  of  the  sides  is 
equal  to  the  product  of  the  other  three,  the  straight  lines  joining  these  points 
with  the  opposite  vertices  of  the  triangle  meet  in  one  point. 

The  proof  is  similar  to  that  of  (3). 

7.  Scholium.  The  principle  of  this  corollary  often  serves  to  determine 
whether  three  straight  lines  meet  in  a  point.     For  example,  if  Aa,  Bb,  Cc, 
are  the  bisectors  of  the  angles  of  the  triangle  ABC,  we  have  by  (III.  21), 

aB=aC[  bC^bA  cA  =  cB 

AB      AC'          BC     AB*  AC      BCr 

and  the  product  of  these  equalities,  omitting  the  common  denominator 

ABXBCX  AC,  is 


therefore,  the  three  bisectors  of  the  angles  of  a  triangle  pass  through  the  same 
point. 

With  the  same  facility,  it  can  be  shown  that  the  straight  lines  joining  the 
points  of  contact  of  the  inscribed  circle  with  the  opposite  vertices  of  the  tri 
angle  meet  in  a  point  ;  that  the  three  perpendiculars  from  the  vertices  of  a 
triangle  to  the  opposite  sides  meet  in  a  point  ;  and  that  the  three  medial  lines 
of  a  triangle  meet  in  a  point. 

29  *  W 


338  MODEEN      GEOMETRY. 


ANHAEMONIC    EATIO. 

8.  Definition.  If  four  points  are  taken  in  a  straight  line,  the  quotient 
obtained  by  dividing  the  ratio  of  the  distances  of  the  first  two  from  the  third 
by  the  ratio  of  the  distances  of  the  first  two  from  the  fourth,  is  called  the 
anharmonic  ratio  of  the  four  points. 

Thus  the  anharmonic  ratio  of  the  four  |  -  \  -  1  -  1 
points  A,  B,  C\  D,  is  A  BCD 

CA  .  DA 

CB  ' 


which  for  brevity  is  denoted  by  [ABCD]. 

In  applying  the  definition  the  points  may  be  taken  in  any  order  we  please, 
but  the  adopted  order  is  always  to  be  indicated  in  the  notation.  Thus,  the 
same  points,  considered  in  the  order  J.,  (7,  B,  D^  give  the  anharmonic  ratio 


9.   The  anharmonic  ratio  of  four  points  is  not  changed  in  value  when  two 
of  the  points  are  interchanged,  provided  the  other  two  are  interchanged  at 
the  same  time. 
Thus 

CA     DA  -  CA.DB 
CB:  DB~CRDA 

\BADW  -  DB  •  CB-CA 
~'         ~  ' 


DA'  CA~  CRDA 
AC.BD 


[DCBA\  = 


AD  '  BD   BC.AD 

BD  AD  =  AC.BD 
BC  '  AC   BC.AD 


Therefore,  [AS CD]  =  [BADC]  =  [CDAB]  =  [DCBA].  There  are 
then  four  different  ways  in  which  the  same  anharmonic  ratio  can  be  ex 
pressed. 

There  are,  in  all,  twenty-four  ways  in  which  the  four  letters  may  be 
written,  and  therefore  four  points  give  rise  to  six  anharmonic  ratios 
differing  in  value.  Three  of  these  six  are  the  reciprocals  of  the  other 
three. 

10.  Definition.  A  system  of  straight  lines  diverging  from  a  point  is  called 
a  pencil;  each  diverging  line  is  called  a  ray;  and  the  point  from  which  they 
diverge  is  called  the  vertex  of  the  pencil. 


MODERN      GEOMETRY. 


339 


PROPOSITION    III.— THEOREM. 

11.  If  a  pencil  of  four  rays  is  cut  by  a  transversal,  any  anharmonic  ratio 
of  the  four  points  of  intersection  is  constant  for  all  positions  of  the  trans 
versal. 

Fig.  1.  Fig.  2. 


Let  0-MNPQ})Q-the  pencil;  and  let  ABCD,  A'B'C'D',  be  any  two 

positions  of  a  transversal ;  then 

[ABCD]  =  [A'B'C'D'}. 
For,  drawing  Bed  parallel  to  OM,  we  have  by  similar  triangles, 


CA 
CB 


OA 


DA 
DB 


OA 
dB 


Dividing  the  first  of  these  equations  by  the  second,  we  have 

[ABCD}  =  ^|- 
Drawing  B'cfd'  parallel  to  0 M,  we  have  in  the  same  manner, 


[A'B'C'D*]  = 


d'B' 
c'B' 


The  second  members  of  these  two  equations  being  equal  (III.  35),  we  have 
[ABCD]  =  [A'B'C'D']. 

It  is  important  to  observe  that  the  preceding  demonstration  applies  when 
the  transversals  cut  one  or  more  of  the  rays  on  opposite  sides  of  the  vertex, 
as  in  Fig.  2. 

12.  Definition.  The  anharmonic  ratio  of  a  pencil  of  four  rays  is  the 
anharmonic  ratio  of  the  four  points  on  these  rays  determined  by  any  trans 
versal.  Thus,  [ABCD],  [AC&D})  etc.,  are  the  anharmonic  ratios  of  the 
pencil  formed  by  the  rays  OM,  ON,  OP,  OQ,  in  the  preceding  figure.  To 


340 


MODERN   GEOMETRY. 


distinguish  the  pencil  in  which  the  ratio  is  considered,  the  letter  at  the 
vertex  is  prefixed  to  the  ratio ;  thus,  [  O.ABCD],  [  O.ACBD],  etc. 

13.  The  angles  of  a  pencil  are  the  six  angles  which  the  rays,  taken  two 
and  two,  form  with  each  other.  It  follows  from  the  preceding  proposition 
that  the  values  of  the  anharmonic  ratios  in  two  pencils  will  be  equal,  if  the 
angles  of  the  pencils  are  equal,  each  to  each. 


14.  Definition.  The  anharmonic  ratio  of  four  fixed 
points  A,  B,  C,  D,  o?i  the  circumference  of  a  circle, 
is  the  anharmonio  ratio  of  the  pencil  formed  by  join 
ing  the  four  points  to  any  variable  point  0  on  the 
circumference. 


PROPOSITION    IV.— THEOEEM. 

15.  The  anharmonic  ratio  of  four  fixed  points  on  the  circumference  of  a 
circle  is  constant. 

For,  the  angles  of  the  pencil  remain  the  same  for  all  positions,  0,  0', 
etc.,  of  its  vertex,  on  the  circumference.  (II.  58.) 

16.  Definition.  If  four  fixed  tangents  to  a  circle  are  cut  by  a  fifth  (vari 
able)  tangent,  the  anharmonic  ratio  of  the  four  points  of  intersection  is 
called  the  anharmonic  ratio  of  the  four  tangents. 


PROPOSITION    V. 
17.   The  anharmonic  ratio  of  four  fixed  tangents  to  a  circle  is  constant. 


For,  let  four  tangents,  touch-      -^\  \     ,  \    / 

ing  the  circle  0  at  the  points      —  -Xr- -^^~^.-4-'— 


A,  B,  C,  D,  be  intersected  by 
any  fifth  tangent  in  M,  N,  P,  Q. 
The  pencil  formed  by  the  rays 
OM,  ON,  OP,  OQ,  will  have 
constant  angles  for  all  positions 
of  the  variable  tangent,  since 
(as  the  reader  can  readily  prove) 
the  angle  MON  will  be  meas 
ured  by  one-half  of  the  fixed  arc  AB,  the  angle  JVOPby  one-half  of  the  arc 
BC,  and  the  angle  POQ  by  one-half  of  the  arc  CD.  The  angles  of  the 
pencil  being  constant,  the  anharmonic  ratio  [O.MNPQ]  is  constant. 

18.  Corollary.  The  anharmonic  ratio  of  four  tangents  to  a  circle  is  equal 
to  the  anharmonic  ratio  of  the  four  points  of  contact. 

For,  if  any  point  0/  in  the  circumference  be  joined  to  A,  B,  C,  D,  the 
pencil  formed  will  have  the  same  angles  as  the  pencil  formed  by  the  rays 
OM,.ON,  OP,  OQ,  since  these  angles  will  also  be  measured  by  one-half 
the  arcs  AB,  BC,  CD,  respectively. 


MODERN      GEOMETRY. 


341 


19.  The  properties  of  anharmonic  ratios  can  be  applied  to  the  demonstra 
tion  of  two  classes  of  theorems,  in  one  of  which  certain  points  are  to  be 
shown  to  lie  in  the  same  straight  line,  and  in  the  other  certain  straight  lines 
are  to  be  shown  to  meet  in  the  same  point.  Corresponding  theorems  in  these 
two  classes  are  placed  side  by  side,  in  the  following  propositions,  in  order  to 
exhibit  their  analogy. 


PKOPOSITION    VI. 


Theorem. 


20.  When  two  pencils  0-ABCD, 
O'-A'B'C'D',  have  the  same  anhar 
monic  ratio  and  a  homologous  ray 
OA  common,  the  intersections  b,  c,  d, 
of  the  other  three  pairs  of  homologous 
rays,  are  in  a  straight  line. 


For,  let  the  straight  line  joining  b 
and  c  meet  OA  in  a,  and  let  the 
points  in  which  it  meets  OD  and 
O'D'  be  called  8  and  d',  respectively. 
By  hypothesis  we  have  (11), 

[abed]  =  [abed'], 

which  can  be  true  only  when  d  and  d' 
coincide ;  but  d  and  d'  being  on  the 
different  lines  OD  and  O'D'  can  co 
incide  only  when  they  are  identical 
with  their  intersection  d.  Therefore, 
a,  b,  c,  d,  are  in  a  straight  line. 


22.   Corollary.  If  one  of  the  anhar 
monic  ratios  of  a  pencil  is  equal  to 
one  of  those  of  a  second  pencil,  the 
remaining  anharmonic  ratios  of  the 
29* 


Theorem. 


21.  When  two  right-lined  figures 
of  four  points  A,  B,  C,  D,  and 
A,  B',  C',  D',  have  the  same  an 
harmonic  ratio  and  a  homologous 
point  A  common,  the  straight  lines 
joining  the  other  three  pairs  of  ho 
mologous  points  meet  in  the  same 
point  0. 


For,  let  0  be  the  point  of  meeting 
of  BB'  and  CC' ;    draw   OA  and 
OD',  and  let  the  point  in  which  OD' 
meets  AD  be  called  d. 
Then  we  have  (11), 

[AB'C'D']  =  [ABC*], 
and,  by  hypothesis, 

[AB'C'D'}  =  [ABCD]; 

hence, 

[ABCd]  =  [ABCD]. 

Therefore  d  coincides  with  D,  and 
the  straight  line  DD'  also  passes 
through  the  point  0. 

23.  Corollary/.  If  one  of  the  an 
harmonic  ratios  of  a  system  of  four 
points  is  equal  to  one  of  those  of  a 
second  system,  the  remaining  anhar- 


342 


MODERN   GEOMETRY. 


first  pencil  are  equal  to  those  of  tlie 
second,  each  to  each. 

For,  let  the  pencils  be  placed  so  as 
to  have  a  common  homologous  ray. 
Since  one  of  the  ratios  has  the  same 
value  in  both  pencils,  the  intersec 
tions  of  the  other  three  pairs  of  ho 
mologous  rays  lie  in  a  straight  line, 
which  is  a  common  transversal ;  and 
then  any  two  corresponding  anhar- 
nionic  ratios  in  the  two  pencils  will  be 
equal  to  that  of  the  four  points  on  the 
common  transversal  (11),  and  there 
fore  equal  to  each  other. 


monic  ratios  of  the  two  systems  are 
equal,  each  to  each. 

For,  let  the  two  systems  be  placed 
so  as  to  have  a  common  homologous 
point.  Since  one  of  the  anharmonic 
ratios  has  the  same  value  in  both  sys 
tems,  the  straight  lines  joining  the 
other  three  pairs  of  homologous  points 
meet  in  a  point;  and  then  any  two 
corresponding  anharmonic  ratios  in 
the  two  systems  are  equal,  being  de 
termined  in  the  same  pencil  (11). 


PROPOSITION    VII. 


Theorem. 


24.  If  two  triangles,  ABC,  A'B'C', 
are  so  situated  that  the  three  straight 
lines,  AA',  BB',  CC/,  joining  their 
corresponding  vertices,  meet  in  a  point, 
0,  the  three  intersections,  a,  b,  c,  of 
their  corresponding  sides,  are  in  a 
straight  line. 


Theorem. 


25.  If  two  triangles,  ABC,  A'B'C', 
are  so  situated  that  the  three  intersec 
tions,  a,  b,  c,  of  their  corresponding 
sides  are  in  a  straight  line,  the  three 
straight  lines,  AA ',  BB',  CC',  join 
ing  their  corresponding  vertices,  meet 
in  a  point,  0. 


For,  let  BA  and  B'A'  meet  OC 
in  D  and  I)',  and  suppose  Oc  to  be 
drawn.  The  pencil  Oc,  OB,  OA, 


For,  let  the  straight  line  abc  meet 
CC'  in  d.  Taking  C  and  C'  as  the 
vertices  of  pencils  having  the  com- 


MODEKN      GEOMETRY. 


343 


0(7,  intersected  by  the  transversals 
cD  and  cD',  gives 

[cBAD]  =  [cB'A'D'}, 

or  considering  these  ratios  as  belong 
ing  to^pencils  whose  vertices  are  C 
and  C/,  respectively, 

[C.cBAD]  =  [C'.cB'A'D']. 

These  pencils  having  a  common  an- 
harmonic  ratio  and  a  common  ray 
CO',  the  intersections  a,  b,  c,  of  the 
other  three  pairs  of  homologous  rays 
are  in  a  straight  line  (20). 


mon  transversal  ac,  we  have,  iden 
tically, 

[C,cdba\  =  [C'.cdba]. 

The  first  pencil  being  cut  by  the  trans 
versal  cBAD,  and  the  second  by  the 
transversal  cB'A'D',  the  preceding 
equation  gives  (11), 

[cDAB]  =  [cD'A'B']. 

The  two  systems,  c,  B,  A,  D,  and 
c,  B',  A',  D',  having  a  common  an- 
harmonic  ratio  and  a  common  ho 
mologous  point  c,  the  lines  BB', 
AA',  DD'  (or  CC'},  meet  in  the 
same  point  0  (21). 


26.  Definition.  Two  triangles  ABC,  A'B'C',  which  satisfy  the  con 
ditions  of  the  preceding  two  theorems,  are  called  liomological ;  the  point  0 
is  called  the  centre  of  homology ;  the  line  abc  is  called  the  axis  ofhomology. 


PEOPOSITION    VIII. 


Theorem. 


27.    In  any  hexagon  ABCDEF 
inscribed  in  a  circle,  the  intersections, 


Theorem. 


28.   In  any  hexagon  ABCDEF 
circumscribed  about  a  circle,  the  three 


344 


MODERN      GEOMETRY. 


L,  M,  JV,  o/  £/ie  ^ree  pairs  of  oppo 
site  sides,  are  in  a  straight  line. 

For,  considering  two  pencils  formed 
by  joining  B  and  F  as  vertices,  with 
A,  C,  D  and  E,  we  have  (15), 
[RACDE]  =  [F.ACDE}. 

Cutting  the  first  pencil  by  the  trans 
versal  LPDE,  and  the  second  pencil 
by  the  transversal  NQDC,  the  pre 
ceding  equation  gives  (11), 

[LPDE]  =  [NCDQ]. 

Since  the  two  systems  of  points 
LPDE  and  NCDQ  have  a  common 
anharmonic  ratio  and  a  common  ho 
mologous  point  D,  the  lines  LN,  PC, 
EQ,  joining  the  other  three  pairs  of 
homologous  points,  meet  in  a  common 
point  M  (21).  Therefore  L,  M,  N, 
are  in  the  same  straight  line. 

This  theorem  is  due  to  PASCAL. 

29.  Corollary  1.  If  the  vertex  D 
is  brought  nearer  and  nearer  to  the 
vertex  (7,  the  side  CD  will  approach 
to  the  tangent  at  C',  therefore,  when 
the  point  D  is  finally  made  to  coin 
cide  with  (7,  the  theorem  will  still 
apply  to  the  resulting  pentagon  if  we 
substitute  the  tangent  at  C  for  the 
side  CD.  The  theorem  then  takes 
the  following  form. 

In  any  pentagon  AB  CEF  inscribed 


diagonals,  AD,  BE,  CF,  joining  the 
opposite  vertices,  intersect  in  the  same 
point. 

For,  regarding  AB,  BC,  CD  and 
EF,  as  fixed  tangents  cut  by  the  tan 
gent  ED  in  P,  N,  D,  E,  and  by  the 
tangent  FA  in  A,  L,  M,  F,  we  have 

(IT), 

[PNDE]  =  [ALMF], 

or,  considering  these  anharmonic  ra 
tios  as  belonging  to  pencils  whose  ver 
tices  are  B  and  C,  respectively, 

[B.PNDE]  =  [C.ALMF}. 
These  two  pencils,  having  a  common 
anharmonic  ratio  and  a  common  ho 
mologous  ray  LN,  the  intersections, 
A,  D,  0,  of  the  other  three  pairs  of 
homologous  rays  are  in  a  straight 
line  (20).  Therefore  AD,  BE  and 
CF,  meet  in  the  same  point  0. 

This  theorem  is  due  to  BRIANCHON. 

30.  Corollary  1.  If  a  vertex  C  is 
brought  into  the  circumference,  the 
sides  BC  and  CD  will  become  a  single 
line  touching  the  circle  at  C.  The 
theorem  will  still  apply  to  the  result 
ing  pentagon  if  we  regard  the  point 
of  contact  of  this  side  as  the  vertex 
of  a  circumscribed  hexagon.  The 
theorem  then  takes  the  following 
form. 

In  any  pentagon  ABDEF  circum 
scribed  about  a  circle,  the  line  joining 


a  vertex  and  the  point  of  contact  of 
the  opposite  side,  and  the  diagonals 
joining  the  other  non-consecutive  ver 
tices  meet  in  a  point. 


MODERN      GEOMETRY. 


345 


in  a  circle,  the  'intersection  N  of  a  side 
with  the  tangent  drawn  at  the  oppo 
site  vertex,  and  the  intersections  L,  M, 
of  the  other  non-consecutive  sides  are 
three  points  in  a  straight  line. 

By  the  same  process  we  can  reduce  the  hexagon  to  a  quadrilateral  and 
finally  to  a  triangle  ;  whence  the  following  corollaries. 


31.    Corollary  IL    In  any  quadri 
lateral  inscribed  in  a  circle,   if  tan 


gents  are  drawn  at  two  consecutive 
vertices,  the  point  of  intersection  of 
each  of  them  with  the  side  passing 
through  the  point  of  contact  of  the 
other,  and  the  intersection  of  the  othe, 
two  sides,  are  three  points  in  a  straight 
line. 

33.   Corollary  III.    In  any  quadri- 


32.  Corollary  IL  In  any  quadri 
lateral  circumscribed  about  a  circle, 
if  we  take  the  points  of  contact  of  two 


adjacent  sides,  and  join  the  point  of 
contact  of  each  side  with  the  vertex  on 
the  other  side,  and  if  the  remaining  two 
vertices  are  joined,  the  three  straight 
lines  so  drawn  meet  in  a  point. 


34.   Corollary  III.  In  any  quadri 


lateral  circumscribed  about  a  circle, 
the  straight  lines  joining  the  points  of 
contact  of  opposite  sides,  and  the  di 
agonals,  are  four  straight  lines  which 
meet  in  a  point. 


346 


MODEKN      GEOMETRY. 


lateral  inscribed  in  a  circle,  the  inter 
sections  of  the  tangents  drawn  at  op 
posite  vertices  and  the  intersections 
of  the  opposite  sides  are  four  points 
in  a  straight  line. 

35.  Corollary  IV.   In  any  triangle 
inscribed  in  a  circle,  the  intersections 


36.  Corollary  IV.  In  any  triangle 
circumscribed  about  a  circle,  the 
straight  lines  joining  the  point  of  con 


tact  of  each  side  with  the  opposite  ver 
tex  meet  in  a  point. 


of  each  side  with  the  tangent  drawn 
at  the  opposite  vertex  are  in  a  straight 
line. 


37.  Scholium.  Pascal's  Theorem  (27)  may  be  applied  to  the 
ABCDEFA,  formed  by  joining  any  six  points  of  the 
circumference  by  consecutive  straight  lines  in  any  order 
whatever,  a  figure  which  may  still  be  called  a  hexagon 
(non-convex),  but  which  for  distinction  has  been  called  a 
hexagram. 

The  demonstration  (27)  applies  to  this  figure,  word  for 
word. 

Brianchon's  Theorem  (28)  may  also  be  extended  to  a 
circumscribed  hexagram,  formed  by  six  tangents  at  any 
six  points  taken  in  any  assumed  order  of  succession. 


figure 


MODERN     GEOMETRY.  347 


HARMONIC    PROPORTION. 


38.  Definition.  Four  points  A,  J%  C,  D,  \ 1 \ 1 

are  called  four  harmonic  points  when  their  ^  c      B  D 

anharmonic   ratio    [ABCD]   is  equal  to 

unity ;  that  is,  when 

CA__DA       *  CA^DA 

CB  '  DB        '          r  CB      DB 

which  agrees  with  the  definition  of  harmonic  points  in  (III.  76). 

39.  Definition.  A  harmonic  pencil  is  a  pencil  of  four  rays  whose  anhar 
monic  ratio  is  equal  to  unity ;  that  is,  a  pencil  0,  which  determines  upon 
any  transversal  a  system  of  four  harmonic  points 

A,  B,  C,  D.  From  (11)  it  follows  that  if  one 
transversal  of  a  pencil  is  divided  harmonically,  all 
other  transversals  of  the  pencil  are  also  divided 
harmonically. 

The  points  A  and  B  are  called  conjugate  points 
with  respect  to  C  and  D,  that  is,  they  divide      /  /     /          \ 

the  distance  CD  harmonically ;    and  C  and  D 

are  called  conjugate  points  with  respect  to  A  and  B,  that  is,  they  divide  the 
distance  AB  harmonically  (III.  76).  In  like  manner,  the  rays  OA  and  OB 
are  called  conjugate  rays  with  respect  to  the  rays  OC  and  OD,  and  are  said 
to  divide  the  angle  COD  harmonically ;  and  the  rays  OC  and  OD  are  con 
jugate  rays  with  respect  to  OA  and  OB,  and  divide  the  angle  A  OB  har- 


PROPOSITION   IX.— THEOREM. 

40.  If  a  straight  line  AB  is  divided  harmonically  at  the  points  C  and  D, 
the  half  of  AB  is  a  mean  proportional  between  the  distances  of  its  middle 
point  0  from  the  conjugate  points  C  and  D;  that  is,  OBZ  =  OC.  OD. 

For,  the  harmonic  proportion, 

CA  :  CB  =  DA  :  DB,  -  J; °      }c     \i J> 

gives,  by  composition  and  division  (III.  10) 
and  (III.  9), 

CA  —CB     CA+  CB      DA  —  DB     DA  +  DB 


or,  OC  :  OB  =  OB:  OD.  [I] 

41.  Corollary.  Conversely,  if  we  have  given  AB  and  its  middle  point  0, 
and  if  C  and  D  are  so  taken  that  ~OB*  =  OC.  OD,  then,  A,  B,  C  and  D 
are  four  harmonic  points. 


348 


MODERN      GEOMETRY. 


For,  the  proportion  [1]  gives 

OB  +  OC  :  OB—OC=OD+OB:  OD—  OB; 


that  is, 


CA  :  CB  = 


:  DB. 


[2] 


42.   Scholium.    The  three  straight  lines  AC,  AB,  AD,  are  in  harmonic 
progression.     For,  the  harmonic  proportion  [2]  may  be  written  thus, 

AC  :  AD  =  AB  —  AC  :  AD  —  AB, 

or,  AC,  AB,  AD,  are  such  that  the  first  is  to  the  third  as  the  difference 
between  the  first  and  second  is  to  the  difference  between  the  second  and 
third ;  that  is,  they  are  in  harmonic  progression,  according  to  the  definition 
commonly  given  in  algebra. 

Of  three  straight  lines  A  C,  AB,  AD,  in  harmonic  progression,  the  second 
AB  is  called  a  harmonic  mean  between  the  extremes  AC  and  AD. 


PEOPOSITION    X.— THEOEEM. 

43.  In  a  complete  quadrilateral,  each  diagonal  is  divided  harmonically  by 
the  other  two. 

Let  ABCDEF  be  a  complete  quadri 
lateral  (4),  and  L,  M,  N,  the  intersections 
of  its  three  diagonals.  In  the  triangle 
AEF,  the  transversal  DBM  gives  (2), 

DF.BA.ME  =  DA.BE.MF, 

M 

and  since  the  three  lines  AL,  FB,  ED,    - 
pass  through  the  common  point  C,  we  have  by  (5), 

DF.BA.LE  =  DA.BE.LF. 
Dividing  one  of  these  equations  by  the  other,  we  have 


ME 
LE 


MF 
LF' 


ME 


LE. 
LF 


therefore,  EF  is  divided  harmonically  at  M  and  L.  Hence,  if  AM  be 
joined,  the  four  rays  AM,  AE,  AL,  AF,  will  form  a  harmonic  pencil ;  con 
sequently  M,  N,  B,  D,  are  also  four  harmonic  points,  or  the  diagonal  BD  is 
divided  harmonically  at  M  and  N.  Finally,  if  FN  be  joined,  the  four  rays 
FM,  FB,  FN,  FD,  will  form  a  harmonic  pencil ;  consequently  L,  N,  C,  A, 
are  four  harmonic  points,  or  the  diagonal  AC  is  divided  harmonically  at  L 
and  A/! 


MODERN      GEOMETRY. 


349 


POLE    AND    POLAK    IN    THE    CIECLE. 

44.  Definition.  If  through  a  fixed  point  Pin  the  plane  of  a  circle  (either 
without  the  circle,  Fig.  1,  or  within  it,  Fig.  2),  we  draw  a  secant  and  deter 
mine  on  this  secant  the  point  Q  the  harmonic  conjugate  of  P  with  respect 
to  the  points  of  intersection  C  and  D,  the  locus  of  Q,  as  the  secant  turns 
about  P,  is  called  the  polar  of  the  point  P,  and  P  is  called  the  pole  of  this 
locus,  with  respect  to  the  circle. 


Fig.  1. 


Fig.  2. 


PROPOSITION    XL— THEOREM. 

45.  The  polar  of  a  given  point  with  respect  to  a  circle  is  a  straight  line 
perpendicular  to  the  diameter  drawn  through  the  given  point. 

Let  P  be  the  given  point  (Figs.  1  and  2),  0  the  centre  of  the  circle,  C 
and  D  the  points  in  which  any  secant  drawn  through  P  cuts  the  circumfer 
ence,  Q  the  harmonic  conjugate  of  P  with  respect  to  C  and  D.  Draw  Q  N 
perpendicular  to  the  diameter  AB  which  passes  through  P.  Draw  DN 
meeting  the  circumference  in  C'.  Join  CN,  DA,  DE. 

Since  PNQ  is  a  right  angle,  the  circumference  described  upon  PQ  as  a 
diameter  passes  through  N:  and  since  CD  is  divided  harmonically  at  P  and 
Q,  the  line  NP  bisects  the  angle  CNC'  (III.  79  and  23) ;  therefore  the 
arcs  A  C  and  A  C/  are  equal.  Hence  the  line  D  A  bisects  the  angle  PDN, 
and  DB,  perpendicular  to  DA,  bisects  the  angle  exterior  to  PDN;  there 
fore  PN  is  divided  harmonically  at  A  and  B  (III.  79),  or  N  is  the  har 
monic  conjugate  of  P  with  respect  to  A  and  B.  Consequently  N  is  a  fixed 
point,  and  Q  is  always  in  the  perpendicular  to  the  diameter  AB,  erected  at 
JV;  that  is,  QN  is  the  polar  of  P. 

46.  Corollary  I.    Hence,  to  construct  the  polar  of  a  given  point  P,  with 
respect  to  a  given  circle,  find  on  the  diameter  AB  drawn  through  Pthe  har 
monic  conjugate  N  of  P  with  respect  to  A  and  B,  and  draw  NQ  perpen 
dicular  to  that  diameter ;  then  NQ  is  the  polar  of  P. 

47.  Corollary  II.    To  find  the  pole  of  a  given  straight  line  NQ,  draw  a 
diameter  AB  perpendicular  to  the  given  line  intersecting  it  in  N,  and  on 

30 


350 


MODERN     GEOMETRY. 


this  diameter  take  P  the  harmonic  conjugate  of  N  with  respect  to  A  and 
B ;  then  P  is  the  pole  of  NQ. 

48.  Corollary  III.    Since  AB  is  divided  harmonically  at  P  and  N,  and 
0^4  =  M£,  we  have  (40), 

OI2=  OP.OJV, 

hence,  the  radius  is  a  mean  proportional  between  the  distances  of  the  polar 
and  its  pole  from  the  centre  of  the  circle. 

This  principle  may  be  used  to  determine  the  point  N  from  P,  or  P  from 
N,  instead  of  the  methods  of  (46)  and  (47). 

49.  Corollary  IV.    When  the  point  P  is  without  the  circle,  its  polar  is  the 
line  TT'  joining  the  points  of  contact  of  the  tangents  drawn  from  P.     For 
the  secant  PCD  turning  about  P  approaches  the  tangent  PT  as  its  limit 
(II.  28) ;  and  at  the  limit,  the  points  C  and  D  and  hence  also  Q  (which  is 
always  between  C  and  D)  all  coincide  with  T.     Therefore  T  and  T/  are 
points  of  the  polar. 

50.  CoroTlai-y  V.   The  polar  of  a  point  on  the  circumference  is  the  tangent 
at  that  point.     For,  as  the  point  P  approaches  the  circumference,  the  point 
N  also  approaches  the  circumference  (since  OP.  ON  =  0J.2) ;   and  when 
OP  becomes  equal  to  OA,  ON  also  becomes  equal  to  OA. 


PROPOSITION  XII.— THEOREM. 

51.  1st.  The  polars  of  all  the  points  of  a  straight  line  pass  through  the 
pole  of  that  line.  2d.  The  poles  of  all  the  straight  lines  which  pass  through 
a  fixed  point  are  situated  on  the  polar  of  that  point. 


Let  XT  be  any  straight  line,  P  its  pole  with  respect  to  the  circle  0,  and 
N'  any  point  of  the  line.  Drawing  OPN,  which  is  perpendicular  to  XY, 
we  have  OP.  ON  =  02*  (48).  Let  PP/  be  drawn  perpendicular  to  ON'; 
then,  the  similar  triangles  OPP',  ONN',  give 

OP'.  ON'  =  OP.  ON  =  OA2, 


therefore,  PP'  is  the  polar  of  N'  (48).     Hence,  1st,  the  polar  of  any  point 
N'  of  the  line  XY  passes  through  the  pole  P  of  that  line ;  2d,  the  pole  P 


MODERN      GEOMETRY. 


351 


of  any  straight  line  XY  which  passes  through  the  point  N'  is  situated  on 
the  polar  PP/  of  that  point. 

52.  Corollary.  The  pole  of  a  straight  line  is  the  intersection  of  the  polars 
of  any  two  of  its  points.  The  polar  of  any  point  is  the  straight  line  joining 
the  poles  of  any  two  straight  lines  passing  through  that  point. 


PROPOSITION    XIII.— THEOREM. 

53.  If  through  a  fixed  point  P,  in  the  plane  of  a  circle,  any  two  secants 
PCD,  PC'D',  are  drawn,  and  their  intersections  with  the  circumference  are 
joined  ly  chords  CG\  DD/,  CD',  C'D,  the  locus  of  the  intersections,  M 
and  N,  of  these  chords,  is  the  polar  of  the  fixed  point  P. 


For,  let  K  and  K'  be  the  points  in  which  CD  and  C/D/  intersect  MK 
Then,  considering  the  complete  quadrilateral  MCC'NDD',  the  systems 
PCKD,  PC'K'D',  are  harmonic  (43) ;  therefore  K  and  K'  are  on  the 
polar  of  P  (44) ;  that  is,  MN  is  the  polar  of  P. 

54.  Corollary  I.   The  secants  NCD*,  NC'D,  being  drawn  through  N, 
the  line  PM  is  the  polar  of  N]  and  in  like  manner  PN  is  the  polar  of  M; 
therefore,  in  any  quadrilateral,  CC'D'D,  inscribed  in  a  circle,  the  intersec 
tion  N  of  the  diagonals  and  the  intersections  M  and  P  of  the  opposite  sides, 
form  a  triangle  MNP  each  vertex  of  which  is  the  pole  of  the  opposite 
side. 

55.  Corollary  II.   As  the  transversal  PC'D'  approaches  to  PCD,  the 
secants  MC,  MD,  approach  to  the  tangents  at  C  and  D  as  their  limits ; 
therefore,  at  the  limit,  the  tangents  at  C  and  D  intersect  on  the  polar  of  P. 
Hence,  if  through  a  fixed  point  P  in  the  plane  of  a  circle  any  secant  PCD 
is  drawn,  and  tangents  CT  and  DT  to  the  circle  are  drawn  at  the  points  of 
intersection,  the  locus  of  the  intersection  T  of  these  tangents  is  the  polar  of  the 
fixed  point  P. 

56.  Corollary  III.   From  the  last  property  it  follows,  that  if  we  draw  tan 
gents  to  the  circle  at  the  vertices  of  the  inscribed  quadrilateral  CC/DD/, 


352 


MODERN   GEOMETRY. 


the  complete  circumscribed  quadrilateral  thus  formed  will  have  for  its  diago 
nals  the  three  indefinite  sides  of  the  triangle  MNP.  Hence,  in  any  complete 
quadrilateral  circumscribed  about  a  circle,  the  three  diagonals  form  a  tri 
angle  each  vertex  of  which  is  the  pole  of  the  opposite  side. 

57.  Corollary  IV.  Combining  (54)  and  (56),  we  arrive  at  the  following 
proposition :  If  at  the  vertices  of  an  inscribed  quadrilateral,  tangents  to  the 
circle  are  drawn  forming  a  circumscnbed  quadrilateral,  then,  1st,  the  interior 
diagonals  of  the  two  quadrilaterals  intersect  in  the  same  point  and  form  a 
harmonic  pencil;  2d,  the  third  diagonals  of  the  completed  quadrilaterals  are 
situated  on  the  same  straight  line,  and  their  extremities  are  four  harmonic 
points. 


RECIPROCAL   POLARS. 

58.  Definition.  From  (51)  it  fol 

lows  that  if  the  points  M,  N,  P,  ft 

are  the  poles  of  the  sides  of  a  poly 

gon  ABCD,  then  the  points  A,  B, 
C,  D,  are  the  poles  of  the  sides  of 

the  polygon  MNPQ.     Each  of  the 

two  polygons  thus  related  is  called 

the  reciprocal  polar  of  the  other, 

with  respect  to  the  circle,  which  re 

ceives  the  name  of  auxiliaiy  circle. 
It  will  be  observed  that  either  of  ^ 
the  two  reciprocal  polars  may  be 
derived  from  the  other  by  either  of 

two  processes.  If  the  polygon  ABCD  is  given  the  Polygon  37  NPQ 
may  be  derived  from  it,  1st,  by  taking  the  poles  M,N,  P,  ft  of  the  sides 
of  the  given  polygon  as  the  vertices  of  the  derived  polygon,  or  2d,  the 
polars  MN,  JVP,  Pft  QM  of  the  vertices  of  the  given  polygon  may  be 
taken  as  the  sides  of  the  derived  polygon.  In  like  Ifa«ner'.1/,the.^lyg°" 
MNPQ  is  given,  the  polygon  ABCD  may  be  derived  .from  it  by  either  o 
these  processes. 

59  Method  of  reciprocal  polars.  Since  the  relation  between  two  recip 
rocal  polars  is  such  that  for  each  line  of  one  figure  there  exists  a  correspond 
ing  point  in  the  other,  and  reciprocally,  any  theorem  m  relation  to  the  hnes 
of  points  of  one  figure  may  be  converted  at  once  into  a  ^orem  ,n  relation 
to  the  points  or  lines  of  the  other.  This  is  called  rec,.procatmg  the  theorem 
The  fecundity  of  this  method  is  especially  proved  m  ite  application  to  the 
theory  of  curves  which  do  not  belong  to  elementary  geometry  ;  but  we  can 
ghe  Lne  simple  illustrations  of  the  nature  of  the  method  by  applying  it  t< 


have  no  difficulty  in  showing  that  the  theorems  which  we 
have  paced  against  each  other,  in  Proposition  VIII.  ,  are  reciprocal  theo 
rems.  Thus,  Sie  reciprocal  polar  of  an  inscribed  hexagon  bemg  the  c,cum- 
scribed  hexagon  formed  by  drawing  tangents  at  the  vertices  of  the  first 


MODERN      GEOMETRY. 


353 


(49),  (50),  we  can  immediately  infer  Brianchon's  Theorem  (28)  from  Pas 
cal's  Theorem  (27) ;  for,  the  diagonals  joining  opposite  vertices  of  the 
circumscribed  hexagon  will  be  the  polars  of  the  intersections  of  opposite 
sides  of  the  inscribed  hexagon  (56),  and  therefore  pass  through  the  pole  of 
the  straight  line  in  which  these  intersections  lie  (51).  Similarly,  the  theorem 
of  Pascal  may  be  directly  inferred  from  that  of  Brianchon. 

The  three  following  propositions  are  of  frequent  use  in  deducing  reciprocal 
theorems. 


PROPOSITION    XIV.— THEOREM. 

60.  The  angle  contained,  by  tivo  straight  lines  is  equal  to  the  angle  con 
tained  by  the  straight  lines  joining  their  poles  to  the  centre  of  the  auxiliary 
circle. 


For,  the  poles  Pand  Q  of  two  straight  lines 
AB  and  CD  are  situated  respectively  on  the 
perpendiculars  let  fall  from  the  centre  of  the 
auxiliary  circle  upon  the  lines  AB  and  CD 
(45). 


PROPOSITION    XV.— THEOREM. 

61.  The  ratio  of  the  distances  of  any  two  points  from  the  centre  of  the 
auxiliary  circle  is  equal  to  the  ratio  of  the  distances  of  each  point  from  the 
polar  of  the  other.  (Salmon's  Theorem. ) 

Let  P  and  Q  be  the  points,  AB  and  CD 
their  polars,  PF  the  distance  of  P  from  CD^ 
and  QE  the  distance  of  Q  from  AB,  and  0 
the  centre  of  the  auxiliary  circle.  Draw 
0PM  and  OQN,  which  will  be  parallel  to 
QE  and  PF  respectively ;  draw  PH  perpen 
dicular  to  OJVand  (^perpendicular  to  OM. 
If  R  is  the  radius  of  the  circle,  we  have 
jR2  =  OP.  OM  =  OQ.  ON (48),  whence 


OP 
OQ 


ON 

OM' 


The  similar  triangles  POH  and  Q  OK  give 


30* 


OP  =  OH. 

OQ       OK 

X 


354  MODERN      GEOMETRY. 

therefore  (III.  12), 

OP      QN  +  OH_  HN  =  PF 
OQ      OM  +  OK      KM      QE 


PROPOSITION    XVI.— THEOEEM. 

62.  The  anharmonic  ratio  of  four  points  in  a  straight  line  is  equal  to  that 
of  the  pencil  formed  ly  the  four  polars  of  these  points. 

For,  the  pencil  formed  by  joining  the  four  points  to  the  centre  of  the 
circle  has  the  same  angles  as  the  pencil  formed  by  their  polars  (60),  and 
these  pencils  have  equal  anharmonic  ratios  (13). 


PROPOSITION    XVII.— PROBLEM. 
63.   It  is  a  known  theorem  that  the  three  perpendiculars  from  the  vertices 

of  a  triangle  to  the  opposite  sides  meet  in  a  point ;  it  is  required  to  determine 

its  reciprocal  theorem  by  the  method  of  reciprocal  polars. 
Let  the    perpendiculars    from    the 

angles  upon  the  opposite  sides  of  the 

triangle  AB  C  meet  in  P.   Let  A'B'  C' 

be  the  reciprocal  polar  triangle  of  AB  C, 

A'  being  the  pole  of  BC,  B'  the  pole 

of  AC,  and  C'  the  pole  of  AB.    The 

pole  of  the  perpendicular  AP  is  a  point 

L  on  the  line  B'C',  since  B'C'  is  the 

polar  of  A  (51) ;  the  pole  of  BP  is  a 

point  Jlf  on  A'C',  and  the  pole  of  CP 

is  a  point  N  on  A'B'.    The  direct 

theorem  being  that  the  three  lines  AP, 

BP,  CP,  meet  in  a  point,  the  recipro 
cal  theorem  will  be  that  their  poles  L, 

M,  N,  are  in  a  straight  line,  the  polar  of  P;   but  we  must  express  the 

reciprocal  theorem  in  relation  to  the  triangle  A'B'C'.     Now  JOW04 
OM,  ON,  and  OA',  OB',  OC',  the  angle  A' OL  is  aright  angle,  by  (60) ; 

and  so  also  B' OM  and   C' ON  are  right  angles.      Hence,  the  reciprocal 
theorem  may  be  expressed  as  follows  :  .  »,  « 

If  from  any  point  0  in  the  plane  of  a  triangle  A'B'  C',  straight  lines 
OA'  OB'  OC',  are  drawn  to  its  vertices,  the  lines  OL,  OM,  ON,  drawn 
at  0 perpendicular  respectively  to  the  lines  OA',  OB',  OC',  meet  the  suit* 
respectively  opposite  to  the  corresponding  vertices  in  three  points,  L,  M,  A, 
which  are  in  a  straight  line. 


MODERN   GEOMETRY. 


355 


RADICAL    AXIS    OF    TWO    CIRCLES. 

64.  Definition.   If  through  a  point  P,  in  the  plane  of  a  circle,  a  straight 
line  is  drawn  cutting  the  circumference  in  the  points  A.  and  B,  the  product 
of  the  segments  into  which  the  chord  AB  is  divided  at  P,  namely,  the 
product  PA.  PB,  is  constant ;  that  is,  independent  of  the  direction  of  the 
secant  (III.  61).    This  constant  product, 

depending  upon  the  position  of  the  point 
jPwith  respect  to  the  circle,  is  called  the 
power  of  the  point  with  respect  to  the 
circle. 

If  we  consider  especially  the  secant 
PCD,  drawn  through  P  and  the  centre 
0  of  the  circle,  and  designate  the  dis 
tance  PO  by  d  and  the  radius  of  the 
circle  by  r,  we  have,  when  the  point  P 
is  without  the  circle,  PC  =  d  —  r,  PD  =  d  +  r,  and  hence  the  power  of 
the  point  is  expressed  by  the  product  (d  —  r]  (d  -f  r]  or  d2  —  r2. 

If  the  point  P  is  within  the  circle,  the  absolute  values  of  PC  and  PD 
are  r  —  d  and  r  -f-  d ;  but  the  segments  PC  and  PD  lying  in  opposite 
directions  with  respect  to  P,  are  conceived  to  have  opposite  algebraic  signs, 
so  that  the  product  PC.  PD  must  be  negative  ;  hence  the  power  of  the  point 
Pis  expressed  by  the  product  —  (r  —  d)  (r  +  d)  =  —  (r2  —  d2)  =  d2  —  r2. 
Thus,  in  all  cases,  whether  the  point  is  without  or  within  the  circle,,  its  power 
is  expressed  by  the  square  of  its  distance  from  the  centre  diminished  l>y  the 
square  of  the  radius. 

65.  When  the  point  P  is  without  the  circle,  its  power  is  equal  to  the  square 
of  the  tangent  to  the  circle  drawn  from  that  point. 

When  the  point  is  on  the  circumference,  its  power  is  zero. 


PROPOSITION    XVIII.— THEOREM. 

66.  The  locus  of  all  the  points  whose  powers  with  respect  to  two  given 
circles  are  equal,  is  a  straight  line  perpendicular  to  the  line  joining  the  centres 
of  the  circles,  and  dividing  this  line  so  that  the  difference  of  the  squares  of 
the  two  segments  is  equal  to  the  difference  of  the  squares  of  the  radii. 

Let  0  and  0/  be  the  centres  of  the  two  circles  whose  radii  are  r  and  r7 ; 
let  P  be  any  point  whose  distances  from  0  and  0/  are  d  and  d/,  then  the 
powers  of  P  with  respect  to  the  two  circles  are  d2  —  r2  and  d/  2  —  r' 2,  and 
these  being  equal,  by  hypothesis,  we  have  d*  —  r2  =  d/  2  —  r' 2,  whence 
d2  —  d/  2  =  r2  —  r' 2.  Now,  drawing  PX  perpendicular  to  0  0',  we  have 
from  the  right  triangles  POX,  PO'X, 


OX*  — 


=P02  —  PO'*  =  d2—d/2=r2—r'2; 


356  MODERN     GEOMETRY. 

therefore,  the  quantity  OXZ  —  0* ' X* ',  being  equal  to  r2  — r/2,  is  constant, 
and  X  is  a  fixed  point.     Hence  the  point  P  is  always  in  the  perpendicular 


Fig.  1. 


S' 


to  00'  erected  at  the  fixed  point  X\  that  is,  this  perpendicular  is  the  locus 
of  P. 

67.  Definition.  The  locus,  PX,  of  the  points  whose  powers  with  respect  to 
two  given  circles  0  and  0'  are  equal,  is  called  the  radical  axis  of  the  two 
circles. 

68.  CoroUary  I.   If  the  two  circles  have  no  point  in  common,  the  radical 
axis  does  not  intersect  either  of  them.  Fig.  1. 

If  the  circles  intersect,  the  power  of  each  of  the  points  of  intersection  is 
equal  to  zero ;  therefore,  each  of  these  points  is  a  point  in  the  radical  axis ; 
hence,  in  the  case  of  two  intersecting  circles,  their  common  chord  is  their 
radical  axis.  Fig.  2. 

If  the  circles  touch  each  other,  either  externally  or  internally,  their  common 
tangent  at  the  point  of  contact  is  their  radical  axis. 

69.  Corollary  II.    From  (65)  and  (67)  it  follows  that  the  tangents  PT, 
PT/,  drawn  to  the  two  circles  from  any  point  of  the  radical  axis  without  the 
circles,  are  equal. 

Hence,  if  SS/  is  a  common  tangent  to  the  two  circles,  intersecting  the 
radical  axis  in  N,  we  have  NS  =  NS'.  Therefore,  the  radical  axis  can  be 
constructed  by  joining  the  middle  points  of  any  two  common  tangents. 


PROPOSITION   XIX.— THEOREM. 

70.  The  radical  axes  of  a  system  of  three  circles,  taken  two  and  two,  meet 
in  a  point. 

Let  0,  0',  0",  be  the  given  circles.  Designate  the  radical  axis  of  0' 
and  0"  by  X,  that  of  0  and  0"  by  X',  and  that  of  0  and  0/  by  X". 
The  three  centres  not  being  in  the  same  straight  line,  the  axes  X  and  X', 
perpendicular  to  the  intersecting  lines  00"  and  0"0',  will  meet  in  a 


MODERN      GEOMETRY. 


357 


certain  point  V.    This  point  will  have  equal  powers  with  respect  to  0/  and 
0",  and  with  respect  to  0  and  0",  consequently  it  will  also  have  equal 


powers  with  respect  to  0  and  0/,  and  is  therefore  a  point  in  their  radical 
axis  X". 

71.  Definition.   The  point  in  which  the  radical  axes  of  a  system  of  three 
circles  meet  is  called  the  radical  centre  of  the  system. 

If  the  three  centres  of  the  circles  are  in  a  straight  line,  the  three  axes  are 
parallel,  and  the  radical  centre  is  at  an  infinite  distance. 

72.  Definition.   Two  circles  0  and  0'  intersect 
orthogonally,  that  is,  at  right  angles,  when  then* 
tangents  at  the  point  of  intersection  are  at  right 
angles,  or,  which  is  the  same  thing,  when  their 
radii,  OT,  0/2]  drawn  to  the  common  point,  are 
at  right  angles. 

Denoting  0  0/  by  d,  and  the  radii  by  r  and  r^, 

we  have  in  the  right  triangle  OTO/,  d2  —  r2  =  r/2 ;  hence,  when  two  cir 
cles  intersect  orthogonally,  the  square  of  the  radius  of  either  is  equal  to  the 
power  of  its  centre  with  respect  to  the  other  circle. 


PROPOSITION  XX.— THEOREM. 

73.  The  radical  axis  of  two  given  circles  is  the  locus  of  the  centres  of  a 
system  of  circles  which  intersect  both  the  given  circles  orthogonally ;  and  the 
line  joining  the  centres  of  the  given  circles  is  the  common  radical  axis  of  all 
the  circles  of  that  system. 

Let  P  be  the  centre  of  any  circle  which  cuts  the  two  given  circles  0  and 
0'  orthogonally;  then,  by  (72),  the  powers  of  the  point  P  with  respect  to 
the  two  circles  are  each  equal  to  the  square  of  the  radius  of  the  circle  P, 
that  is,  equal  to  each  other ;  therefore,  the  centre  P  is  in  the  radical  axis  of 
the  two  given  circles. 

Again,  let  P  and  Q  be  the  centres  of  any  two  of  the  circles  which  cut  both 


358 


MODEEN     OEOMETEY. 


0  and  0/  orthogonally.     Since  the  circle  0  cuts  the  two  circles  P  and  Q 
orthogonally,  its  centre  0  lies  in  the  radical  axis  of  P  and  Q ;  and  for  the 


Fig.  1. 


Fig.  2. 


same  reason  the  centre  0/  lies  in  the  radical  axis  of  P  and  Q;  therefore,  the 
line  0  0'  is  that  radical  axil*  and  is  consequently  the  common  radical  axis 
of  all  the  circles  which  cut  both  0  and  0/  orthogonally. 

74.  Scholium.  When  the  given  circles  intersect,  Fig.  1,  the  radius  of  any 
one  of  the  circles  P,  Q,  etc.,  is  evidently  less  than  the  distance  of  its  centre 
from  0  0',  and  therefore  no  one  of  these  circles  cuts  0  0' . 

But  when  the  circles  have  no  point  in  common,  Fig.  2  (whether  one  circle 
is  wholly  without  the  other,  as  in  Fig.  2,  or  wholly  within  the  other),  all  the 
circles,  P,  Q,  etc. ,  cut  the  line  0  0' ;  and  since  0  0'  is  their  common  radical 
axis,  it  is.  their  common  chord ;  therefore,  these  circles  all  pass  through  two 
fixed  points  L  and  L'  in  the  line  0  0/. 

Also,  since  OT  is  a  tangent  to  the  circle  P,  we  have  OL.  OL'  —  ~OT2  = 
OB2 ;  therefore,  the  diameter  AB  is  divided  harmonically  at  L  and  L'  (41). 
For  a  like  reason,  A/B/  is  divided  harmonically  at  L  and  L'. 


CENTEES    OF    SIMILITUDE    OF    TWO    CIECLES. 

75.  Definition.  If  the  straight  line  joining  the  centres  of  two  circles  is 
divided  externally  and  internally  in  the  ratio  of  the  corresponding  radii,  the 
points  of  section  are  called,  respectively,  the  external  and  the  internal  centres 
of  similitude  of  the  two  circles. 


PROPOSITION    XXI.— THEOREM. 

76.  If  in  two  circles  two  parallel  radii  are  drawn,  one  in  each  circle,  the 
straight  line  joining  their  extremities  intersects  the  line  of  centres  in  the  exter 
nal  centre  of  similitude  if  the  parallel  radii  are  in  the  same  direction,  and  in 
the  internal  centre  of  similitude  if  these  radii  are  in  opposite  directions. 


MODERN      GEOMETRY.  359 

For,  OA  and  O'A'  being  any  two  parallel  radii  in  the  same  direction,  and 


the  line  A' A  intersecting  the  line  of  centres  in  S,  the  similar  triangles 
SOA,  SO' A',  give 

SO  :  SO'  =  OA  :  O'A', 

and  therefore,  by  the  definition  (75),  Sis  the  external  centre  of  similitude. 
Also,  OA  and  O'Ai  being  parallel  radii  in  opposite  directions,  and  the 
line  AAl  intersecting  the  line  of  centres  in  T,  the  similar  triangles  TO  A, 
TO' A^  give 

TO  :  TO'  =  OA  :  O'A,, 

and  therefore  T  is  the  internal  centre  of  similitude. 

77.  Corollary  I.    It  is  easily  shown  that,  conversely,  if  any  transversal  is 
drawn  through  a  centre  of  similitude,  the  radii  drawn  to  the  points  in  which 
it  cuts  the  circumferences  will  be  parallel,  two  and  two. 

Of  the  four  points  in  which  the  transversal  cuts  the  circumferences,  two 
points  at  the  extremities  of  parallel  radii,  as  A  and  A',  or  B  and  B',  are 
called  homologous  points;  and  two  points  at  the  extremities  of  non-parallel 
radii,  as  A  and  B',  or  B  and  A',  are  called  anti-homologous  points. 

78.  Corollary  II.    Hence,  if  a  transversal  drawn  through  a  centre  of 
similitude  is  a  tangent  to  one  of  the  circles  it  is  also  a  tangent  to  the  other ; 
so  that  when  one  circle  is  wholly  without  the  other,  the  centres  of  similitude 
are  the  intersections  of  the  pairs  of  external  and  internal  common  tangents, 
respectively. 

Fig.  1. 

Fig.  2. 
A' 


If  the  circles  touch  each  other  externally  (Fig.  1),  the  point  of  contact  is 


360 


MODERN   GEOMETRY. 


their  internal  centre  of  similitude.     If  they  touch  internally  (Fig.  2),  the 
point  of  contact  is  their  external  centre  of  similitude. 

Fig.  3. 


If  one  circle  is  wholly  within  the  other  (Fig.  3), 
both  centres  of  similitude  lie  within  both  circles. 


79.  Corollary  III,   The  distances  (as  SA  and  SA',  or  TA  and  TA^  etc.) 
of  a  centre  of  similitude  from  two  homologous  points  are  to  each  other  as 
the  radii  of  the  circles. 

80.  Corollary  IV.  Since  we  have 

SO  :  SO'^TO  :  TO', 

the  line  00'  is  divided  harmonically  at  $  and  T;  that  is,  the  centres  of 
two  circles  and  their  two  centres  of  similitude  are  four  harmonic  points. 


PKOPOSITION    XXII.— THEOEEM. 

81.    The  product  of  the  distances  of  a  centre  of  similitude  of  two  circles 
from  two  anti-homologous  points  is  constant. 


Let  a  transversal  through  the  centre  of  similitude  S  intersect  the  circum 
ferences  0  and  0'  in  the  homologous  points  A.,  A',  and  B,  B'.  The  line 
of  centres  intersects  the  circumferences  in  the  homologous  points  M,  M', 
and  N,  N',  respectively.  Hence,  by  (79), 


ON 
O'N' 


SA 

SA' 


SB 
SB' 


SM        SN  . 
SM'      SN' ' 


MODEKN     GEOMETRY. 

from  which  equations  we  deduce 

SA.  SB'  =  SA'.  SB  =  J^t,  X  SA'.  SB'. 

But  SA'.  SB'  =  SM'.  SN'  (III.  58) ;  therefore  we  have 

SA.  SB'  =  SA'.  SB  =  SM.  SN'  =  SM'.  SN. 

The  products  SM.  SN',  SM'.  SN,  are  constant ;  therefore,  the  products 
SA.SB',  SA'.SB,  are  constant. 

82.  Corollary  I.   Hence,  if  A  and  B'  are  anti-homologous  points  /of  one 
secant  drawn  through   S,  and  b  and  a'  are  anti-homologous  points  of  a 
second  secant,  we  have 

SA.SB'  =  Sb.Sa'; 

therefore,   the  four  points  A,   B',  a',  b,  lie  on  the  circumference  of  a 
circle  0". 

83.  Corollary  II.   The  chords  Ab,  a'B',  joining  pairs  of  anti-homologous 
points  in  the  two  given  circles,  may  be  called  anti-homologous  chords. 

The  chord  Ab  is  the  radical  axis  of  the  circles  0  and  0" ;  the  chord  a/ B' 
is  the  radical  axis  of  the  circles  0'  and  0"  (68) ;  and  these  intersect  the 
radical  axis  PXof  the  circles  0  and  0'  in  the  same  point  P  (70).  Hence, 
pairs  of  anti-homologous  chords  in  two  circles  intersect  on  the  radical  axis 
of  the  circles. 

84.  Corollary  III.    If  the  secant  Sa/  approaches  indefinitely  to  SA',  the 
anti-homologous  chords  a' A',  bB,  approach  indefinitely  to  the  tangents  at 
A'  and  B.     Hence,  at  the  limit,  we  infer  that  the  tangents  at  two  anti- 
homologous  points  in  two  circles  intersect  on  the  radical  axis. 


PROPOSITION    XXIII.— THEOREM. 

85.  Three  circles  being  given,  and  considered  when  taken  two  and  two  as 
forming  three  pairs  of  circles',  then,  1st.  The  straight  lines  joining  the 
centre  of  each  circle  and  the  internal  centre  of  similitude  of  the  other  two 
meet  in  a  point;  2d.  The  external  centres  of  similitude  of  the  three  pairs 
of  circles  are  in  a,  straight  line  ;  3d.  The  external  centre  of  similitude  of 
any  pair  and  the  internal  centres  of  similitude  of  the  other  two  pairs  are 
in  a  straight  line. 

Let  0,  0',  0",  be  the  given  circles ;   S  and  T  the  external  and  internal 
centres  of  similitude  of  0'  and  0" ;  S'  and  T'  those  of  0  and  0" ;  S" 
and  T"  those  of  0  and  0'.     Let  R,  R'  and  R"  denote  the  radii  of  the 
three  circles. 
31 


362 


MODERN      GEOMETRY. 


1st.  By  the  definition  (75)  we  have 

T"0  _R  TO'  _R' 

T"0'~R''        TO"~~R" 

the  product  of  which  equations  gives 


T'O 

T'O 


or 


T"O.TO'.T'Q"  =  R.R'.R"  _ 
T" 0'. TO". T'0~  R'.R".R  ~   ' 

T"  0.  TO'.  T'  0"  =  T"  0/.  TO".  T'  0 ; 


therefore,  in  the  triangle   00/0//,  the  three  straight  lines  OT,  O'T', 
0"T",  meet  in  a  point  (6). 
2d.  By  the  definition  we  also  have 


S"0       R  SO'       R/ 


S'O"      R' 


S"0'      R'          SO"      R"         S'O        R 

whence,  by  multiplying  these  equations  together, 

S"  0.  SO'.  S'  0"  =  S"  0'.  SO".  S'  0 ; 


MODEEN      GEOMETKY.  363 

therefore,  the  points  S,  S',  £",  being  in  the  sides  (produced)  of  the  tri 
angle  OO'O",  are  in  a  straight  line  (3). 

3d.  The  product  of  the  equations 

T"0  =  R_         TO'       JV_         S'O"  _  R" 
T" 0'      R''        TO"  ~  R" '        WO~  ~  W ' 

gives  T"  0.  TO'.  S'  0"  =  T"  0'.  TO".  S'  0 ; 

therefore,  the  points  T,  T",  S',  are  in  a  straight  line  (3).  In  the  same 
manner  it  is  shown  that  T,  T',  S",  are  in  a  straight  line ;  and  T'  T"  S 
are  in  a  straight  line. 

86.  Definition.  The  straight  line  SS'S",  on  which  the  three  external 
centres  of  similitude  lie,  is  called  the  external  axis  of  similitude  of  the  three 
circles ;  and  the  lines  ST"T',  S'T"T,  S"T'T,  are  called  the  three  inter 
nal  axes  of  similitude. 


PROPOSITION    XXIV.— THEOREM. 

87.  If  a  variable  circle  touch  two  fixed  circles,  the  chord  of  contact  passes 
through  their  external  centre  of  similitude  when  the  contacts  are  of  the  same 
kind  (both  external  or  loth  internal],  and  through  their  internal  centre  of 
similitude  when  the  contacts  are  of  different  kinds. 

Let  the  circle  C  touch  the  circles  0  and  0'  in  A  and  B' ;  and  let  the 
chord  of  contact  AB'  cut  the  two  circles  again  in  B  and  A'.  The  lines  OC, 
O'C,  pass  through  the  points  of  contact.  Drawing  the  radii  0£,  0'A'\ 


Fig.  1. 


the  isosceles  triangles  CAB',  OAB,  O'A'B',  are  similar ;  consequently  the 
radii  OA  and  O'A'  are  parallel.  Therefore  (76),  the  chord  AB'  passes 
through  the  external  centre  of  similitude  S,  when  the  contacts  are  of  the 


364  MODERN     GEOMETRY. 

same  kind  (Fig.  1),  and  through  the  internal  centre  of  similitude  T:  when 
the  contacts  are  of  a  different  kind  (Fig.  2). 

Fig.  2. 


88.  Corollary.    If  two  variable  circles  C  and  c  touch  two  fixed  circles  0 
and  0',  their  radical  axis  passes  through  the  external  centre  of  similitude 
of  the  fixed  circles  when  the  contacts  of  each  of  the  two  circles  are  of  the  same 
kind,  and  through  their  internal  centre  of  similitude  when  these  contacts  are 
of  different  lands.  4 

For,  the  four  points  of  contact  A,  W,  I',  a,  (Fig.  1),  lie  on  the  circumfer 
ence  of  a  circle  (82)  which  may  be  designated  as  the  circle  Q.  The  chord 
AB/  is  the  radical  axis  of  the  circle  Q  and  the  circle  C\  the  chord  ab'  is 
the  radical  axis  of  the  circle  Q  and  the  circle  c ;  and  these  two  axes  meet  the 
radical  axis  of  the  circles  C  and  c  in  the  same  point  (70),  that  is,  in  the 
point  S  (87).  The  proof  is  similar  when  the  contacts  are  of  different  kinds. 

PROPOSITION    XXV.— THEOREM. 

89.  The  radical  axis  of  two  circles  which  touch  three  given  circles  is  an 
axis  of  similitude  of  the  three  given  circles. 

Let  the  circles  M  and  N  (figure  on  next  page)  touch  the  three  given 
circles  0,  0',  0",  the  contacts  of  each  of  the  two  circles  being  all  of  the 
same  kind,  that  is,  all  internal  in  the  case  of  the  circle  M,  and  all  external 
in  the  case  of  the  circle  N.  Let  8,  S',  S",  be  the  three  external  centres 
of  similitude  of  the  given  circles  taken  in  pairs,  so  that  SS'S"  is  their 
external  axis  of  similitude  (86). 

Since  the  circles  M  and  N  touch  the  two  given  circles  0'  and  0",  and 
the  contacts  are  of  the  same  kind  in  each  case,  the  radical  axis  of  Jf  and  N 
passes  through  S  (88).  For  the  same  reason,  it  passes  through  S'  and 
through  S".  Therefore  SB'S"  is  the  radical  axis  of  the  circles  M  and  N. 

In  the  same  manner  it  may  be  shown  that  if  each  of  the  two  circles  M 
and  N  has  like  contacts  with  the  pair  of  circles   0'  and  0",  but  unlike 
contacts  with  the  other  two  pairs  (that  is,  if  M  touches  both  0'  and  0/ 
internally  and  0  externally,  and  N  touches  both  0/  and  0"  externally  and 
0  internally),  the  radical  axis  of  M  and  N  is  the  internal  axis  of  similitude 
which  passes  through  the  external  centre  of  similitude,  S,  of  the  circles  0 
and  0". 


MODERN      GEOMETRY. 


365 


90.  Scholium.   There  are  in  general  eight  different  circles  which  can  be 
drawn  tangent  to  three  given  circles,  and  these  eight  circles  exist  in  pairs 
the  four  radical  axes  of  which  are  the  four  axes  of  similitude  of  the  three 
given  circles. 

91.  Corollary  I.    When  two  circles  M  and  N  touch  three  given  circles 
0,  #',  0",  the  three  chords  of  contact  mn,  m'n',  m"n",  meet  in  a  point 
V,  which  is  a  centre  of  similitude  of  the  two  circles  M  and  N  and  the  radical 
centre  of  the  three  given  circles  0,  0',  0". 

For,  since  the  circle  0  touches  the  circles  M  and  N,  and  the  contacts  are 
of  different  kinds,  the  chord  of  contact  mn  passes  through  the  internal  centre 
31* 


366  MODERN      GEOMETRY. 

of  similitude  V  of  M  and  N  (87) ;  and  for  the  same  reason  the  chords 
m'n'  and  m//n//  pass  through  V. 

Also,  since  the  two  circles  0  and  0/  touch  the  two  circles  M  and  JVJ  and 
the  contacts  are  of  different  kinds,  the  radical  axis  of  0  and  0/  passes 
through  the  internal  centre  of  similitude,  F,  of  M  and  N  (88).  For  the 
same  reason,  the  radical  axis  of  0'  and  0",  and  the  radical  axis  of  0"  and 
0',  pass  through  V.  Therefore,  V  is  the  radical  centre  of  the  three  circles 
0,  0',  0". 

92.  Corollary  II.  The  pole  of  the  radical  axis  of  the  circles  M  and  N, 
with  reference  to  any  one  of  the  three  given  circles,  lies  in  the  chord  of  contact 
of  that  circle.  Thus,  in  the  case  represented  in  the  figure,  the  pole  of 
SS'S"  with  respect  to  the  circle  0  is  a  point  P  lying  in  the  chord  of 
contact  mn. 

For,  let  R  be  the  point  of  meeting  of  .the  tangents  to  the  circle  0  drawn 
at  m  and  n.  These  tangents  are  equal  and  touch  the  circles  M  and  N, 
therefore,  the  point  R  is  on  the  radical  axis  of  M  and  JV,  that  is,  upon  the 
line  SS'S".  But  mn  is  the  polar  of  the  point  R  with  respect  to  the  circle 
0  (49),  and  therefore  the  pole  of  SS' S"  with  respect  to  the  circle  0  is  a 
point  P  on  the  chord  of  contact  mn  (51). 


PROPOSITION    XXVI.— PROBLEM. 

93.    To  describe  a  circle  tangent  to  three  given  circles. 

As  remarked  in  (90),  there  are  in  general  eight  solutions  of  this  problem. 
The  solutions  may  all  be  brought  under  two  cases :  viz. — 

1st.  A  pair  of  circles  can  be  found  one  of  which  will  touch  all  the  given 
circles  internally,  and  the  other  will  touch  all  the  given  circles  externally. 

2d.  A  pair  of  circles  can  be  found  one  of  which  will  touch  the  first  of  the 
given  circles  internally  and  the  other  two  externally,  and  the  other  will  touch 
the  first  externally  and  the  other  two  internally. 

By  taking  each  of  the  given  circles  successively  as  the  "firsfo"  this  second 
case  gives  six  circles,  thus  making,  in  all,  the  eight  solutions. 

The  principles  developed  in  the  preceding  proposition  furnish  the  follow 
ing  simple  and  elegant  solution  of  the  problem,  first  given  by  GTERGONNE.* 

Let  0,  0',  0"  (preceding  figure)  be  the  three  given  circles.  Let  SS'S" 
be  their  external  axis  of  similitude  and  V  their  radical  centre.  Find  the 
poles  P,  P',  P",  of  SS'S"  with  respect  to  each  of  the  given  circles,  and 
draw  FP,  VP',  VP",  intersecting  the  three  circles  in  the  points  m  arid  n, 
mf  and  n',  m"  and  n",  respectively.  The  circumference  described  through 
the  three  points  m,  m',  m",  will  touch  the  three  given  circles  internally; 
and  the  circumference  described  through  the  three  points  n,  n',  n",  will 
touch  the  three  given  circles  externally. 

By  substituting  successively  each  internal  axis  of  similitude  for  SS'S", 
we  obtain  the  other  three  pairs  of  circles. 

*  Annales  de  Mathtmatiques,  t.  IV. 


MODERN      GEOMETRY.  367 

94.  Scholium.  This  general  solution  embraces  the  solution  of  ten  distinct 
problems,  special  cases  of  the  general  problem,  in  which  one  or  more  of  the 
given  circles  may  be  reduced  to  points  (that  is,  circles  of  infinitely  small 
radius)  or  to  straight  lines  (that  is,  circles  of  infinitely  great  radius). 


EXERCISES. 

1.  If  L  and  I/  are  two  fixed  straight  lines  and  0  a  fixed  point,  and  if 
through  0  any  two  straight  lines  OAA',  OBB',  are  drawn  cutting  L  in  A 
and  E  and  U  in  A'  and  B^  find  the  locus  of  the  intersection  of  the  lines 
AB'  and  AfB  (43). 

2.  If  the  three  sides  of  a  triangle  pass  through  three  fixed  points  which 
are  in  a  straight  line,  and  two  vertices  of  the  triangle  move  on  two  fixed 
straight  lines,  the  third  vertex  moves  on  a  straight  line  which  passes  through 
the  intersection  of  the  two  fixed  lines  (25). 

3.  If  the  three  vertices  of  a  triangle  move  on  three  fixed  straight  lines 
which  meet  in  a  point,  and  two  sides  of  the  triangle  pass  through  two  fixed 
points,  the  third  side  passes  through  a  fixed  point  which  is  in  a  straight  line 
with  the  other  two  (24). 

4.  If  Q  is  any  point  in  the  polar  of  a  point  P  with  respect  to  a  given 
circle,  the  circle  described  upon  PQ  as  a  diameter  cuts  the  given  circle 
orthogonally  (48). 

5.  Let  the  polars  of  any  point  P,  with  respect  to  two  given  circles  0  and 
0',  intersect  in  Q.     Then,  the  circle  described  upon  PQ  as  a  diameter  cuts 
both  the  given  circles  orthogonally,  and  its  centre  is  on  the  radical  axis  of 
the  given  circles. 

6.  Describe  a  circumference  which  shall  pass  through  a  given  point  and 
cut  two  given  circles  orthogonally. 

7.  The  polars  of  any  point  in  the  radical  axis  of  two  circles  intersect  on 
that  axis. 

8.  The  poles  of  the  radical  axis  of  two  circles  taken  with  respect  to  each 
circle,  and  the  two  centres  of  similitude  of  the  circles,  are  four  harmonic 
points. 

9.  The  radical  axis  of  two  circles  is  equally  distant  from  the  two  polars 
of  either  centre  of  similitude. 

10.  If  the  sides  AB,  BC,   CD,  DA,  of  a  quadrilateral  circumscribed 
about  a  circle  whose  centre  is   0  touch  the  circumference  at  the  points 
E,  F,  G,  H,  respectively,  and  if  the  chords  HE  and  GF  meet  in  P,  the 
line  PO  is  perpendicular  to  the  diagonal  A  C. 

11.  If  a  quadrilateral  is  divided  into  two  other  quadrilaterals  by  any 
secant,  the  intersections  of  the  diagonals  in  the  three  quadrilaterals  are  in  a 
straight  line. 


368  MODEEN     GEOMETRY. 

12.  The  anharmonic  ratio  of  four  points  on  the  circumference  of  a  circle  is 
equal  to  the  ratio  of  the  products  of  the  opposite  sides  of  the  quadrilateral 
determined  by  these  points. 

13.  If  a  series  of  circles  having  their  centres  in  a  given  straight  line  cut  a 
given  circle  orthogonally,  they  have  a  common  radical  axis,  which  is  the 
perpendicular  let  fall  from  the  centre  of  the  given  circle  upon  the  given 
straight  line. 

14.  The  three  circles  described  upon  the  diagonals  of  a  complete  quadri 
lateral  as  diameters  have  a  common  radical  axis  and  cut  orthogonally  the 
circle  described  about  the  triangle  formed  by  the  three  diagonals. 

15.  Three  circles  01;  02,  03,  being  given,  any  fourth  circle  Q  is  described 
and  the  radical  axes  of  Q  and  each  of  the  given  circles  are  drawn  forming  a 
triangle  ABC.     Another  circle  Q'  being  drawn,  a  second  triangle  A'B'C' 
is  formed  in  the  same  manner.    Prove  that  the  triangles  AB  C  and  A/B/  Cf 
are  homological  (24),  (70). 

16.  If  two  triangles  are  reciprocal  polars  with  respect  to  a  circle,  they  are 
homological  (51),  (62),  (20). 

17.  If  from  the  vertices  of  a  triangle  ABC  perpendiculars  Aa,  Bl,  Cc, 
are  let  fall  upon  the  opposite  sides,  the  three  pairs  of  sides  BC  and  6c, 
AC  and  ac,  AB  and  a&,  intersect  on  the  radical  axis  of  the  circles  circum 
scribed  about  the  triangles  ABC  and  abc  (64,  67). 

18.  Any  common  tangent  to  two  circles  is  divided  harmonically  by  any 
circle  which  has  a  common  radical  axis  with  the  two  given  circles  (41). 

19.  If  the  sides  of  a  quadrilateral  ABCD  inscribed  in  a  circle  are  pro 
duced  to  meet  in  E  and  F,  forming  a  complete  quadrilateral,  the  square  of 
the  third  diagonal  EF  is  equal  to  the  sum  of  the  squares  of  the  tangents 
from  E  and  F;  and  the  tangent  from  the  middle  point  of  .E^is  equal  to 
one-half  of  EF. 

20.  Given  the  three  diagonals  of  a  complete  quadrilateral  inscribed  in  a 
given  circle,  it  is  required  to  construct  the  quadrilateral  (4,  48,  49,  57). 

21.  Given  three  circles,  it  is  required  to  describe  a  fourth  such  that  the 
three  radical  axes  of  this  circle,  combined  successively  with  each  of  the 
given  ones,  shall  pass  through  three  given  points  (Exercises  3  and  15). 


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